Recent questions in Comparing two groups

College StatisticsAnswered question

Monique Henson 2023-04-01

The product of the ages, in years, of three (3) teenagers os 4590. None of the have the sane age. What are the ages of the teenagers???

College StatisticsAnswered question

akuzativo617 2022-11-10

The difference between vector space and group

When comparing the difference between the definition of vector space, I see that the main job is that vector space defines a scalar product while the group not, so here list two of my questions?

1. Why we need to define a scalar product for a vector space? Physical sense or some insight behind it?

2. One truly nice thing for vector space is that we represent the element with basis, so what we do with elements in vector space is just with basis,so why we can't do the same thing for group?

When comparing the difference between the definition of vector space, I see that the main job is that vector space defines a scalar product while the group not, so here list two of my questions?

1. Why we need to define a scalar product for a vector space? Physical sense or some insight behind it?

2. One truly nice thing for vector space is that we represent the element with basis, so what we do with elements in vector space is just with basis,so why we can't do the same thing for group?

College StatisticsAnswered question

Joglxym 2022-11-10

Argument that two given finite groups are not isomorphic.

Two finite groups are to be compared in terms of structure (whether isomorphic or not): $\u27e8\phantom{\rule{thinmathspace}{0ex}}x,y\mid {x}^{2},\text{}{y}^{2},\text{}(xy{)}^{2}\phantom{\rule{thinmathspace}{0ex}}\u27e9$ and $\u27e8\phantom{\rule{thinmathspace}{0ex}}x,y\mid {x}^{4},\text{}{y}^{2}={x}^{2},\text{}yx{y}^{-1}={x}^{-1}\phantom{\rule{thinmathspace}{0ex}}\u27e9$.

First group is very simple, the second one not so much. However I found both of them on Wikipedia - their orders are 4 and 8 respectively.

So for the second group do I still have to figure out all the relations (to arrive at order 8) just to make a statement that orders are not equal hence no isomorphism? Is there a quick formal argument for example that order is greater than 4?

Two finite groups are to be compared in terms of structure (whether isomorphic or not): $\u27e8\phantom{\rule{thinmathspace}{0ex}}x,y\mid {x}^{2},\text{}{y}^{2},\text{}(xy{)}^{2}\phantom{\rule{thinmathspace}{0ex}}\u27e9$ and $\u27e8\phantom{\rule{thinmathspace}{0ex}}x,y\mid {x}^{4},\text{}{y}^{2}={x}^{2},\text{}yx{y}^{-1}={x}^{-1}\phantom{\rule{thinmathspace}{0ex}}\u27e9$.

First group is very simple, the second one not so much. However I found both of them on Wikipedia - their orders are 4 and 8 respectively.

So for the second group do I still have to figure out all the relations (to arrive at order 8) just to make a statement that orders are not equal hence no isomorphism? Is there a quick formal argument for example that order is greater than 4?

College StatisticsAnswered question

racmanovcf 2022-10-18

How can I compare two series that are same thing, but at different rates?

Consider the series $1-\frac{1}{2}+\frac{2}{3}-\frac{1}{3}+\frac{2}{4}-\frac{1}{4}+\frac{2}{5}-\frac{1}{5}$

I can see that if you group the terms in pairs of two, you get $(1-\frac{1}{2})+(\frac{2}{3}-\frac{1}{3})+(\frac{2}{4}-\frac{1}{4})+(\frac{2}{5}-\frac{1}{5})$...= $\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}$...

The partial sums ${S}_{2n}$= the harmonic series from $$n=2$$ to n.

Consider the series $1-\frac{1}{2}+\frac{2}{3}-\frac{1}{3}+\frac{2}{4}-\frac{1}{4}+\frac{2}{5}-\frac{1}{5}$

I can see that if you group the terms in pairs of two, you get $(1-\frac{1}{2})+(\frac{2}{3}-\frac{1}{3})+(\frac{2}{4}-\frac{1}{4})+(\frac{2}{5}-\frac{1}{5})$...= $\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}$...

The partial sums ${S}_{2n}$= the harmonic series from $$n=2$$ to n.

College StatisticsAnswered question

taumulurtulkyoy 2022-10-16

Analogy between quotient groups and quotient topology

I am trying to see "analogy" between the two concepts in the title, although I am familiar with independent definitions of them.

(1) Let G be a group, N a normal subgroup. Define an equivalence relation on G by ${g}_{1}\sim {g}_{2}$ if ${g}_{1}{g}_{2}^{-1}\in N$. Let $[{g}_{1}]$ denote equivalence class of g, and G/N the set of equivalence classes. Then we have a canonical map $\phi :G\to G/N$, where domain is group, and codomain is "set" right now.

Then the quotient structure on G/N is a "group structure" such that the map φ becomes a homomorphism. Such a structure is unique. We call G/N quotient group.

Step 2

(2) Let X be a topological space, ∼ an equivalence relation on X, and [x] denote equivalence class. Let X/∼ denote the set of equivalence classes. Then there is canonical map $\phi :X\to X/\sim $, where domain is topological space and codomain is just set right now.

The quotient topology on $X/\sim $ is a topology such that the map $\phi $ is continuous.

But, I was feeling that there could be more than one topological structures on X/∼ which make the canonical map continuous. For example, take nice example of quotient topology on one hand, and structure of indiscrete topology on X/∼ on the other hand.

Question: What is exact definition of quotient topology which makes it unique in some sense? (compare with quotient structure in (1))

I am trying to see "analogy" between the two concepts in the title, although I am familiar with independent definitions of them.

(1) Let G be a group, N a normal subgroup. Define an equivalence relation on G by ${g}_{1}\sim {g}_{2}$ if ${g}_{1}{g}_{2}^{-1}\in N$. Let $[{g}_{1}]$ denote equivalence class of g, and G/N the set of equivalence classes. Then we have a canonical map $\phi :G\to G/N$, where domain is group, and codomain is "set" right now.

Then the quotient structure on G/N is a "group structure" such that the map φ becomes a homomorphism. Such a structure is unique. We call G/N quotient group.

Step 2

(2) Let X be a topological space, ∼ an equivalence relation on X, and [x] denote equivalence class. Let X/∼ denote the set of equivalence classes. Then there is canonical map $\phi :X\to X/\sim $, where domain is topological space and codomain is just set right now.

The quotient topology on $X/\sim $ is a topology such that the map $\phi $ is continuous.

But, I was feeling that there could be more than one topological structures on X/∼ which make the canonical map continuous. For example, take nice example of quotient topology on one hand, and structure of indiscrete topology on X/∼ on the other hand.

Question: What is exact definition of quotient topology which makes it unique in some sense? (compare with quotient structure in (1))

College StatisticsAnswered question

Payton Rasmussen 2022-10-09

(1) Prove that O(n) is homeomorphic to $\mathrm{S}\mathrm{O}(n)\times {\mathbb{Z}}_{2}$. (2) Are these two isomorphic as topological groups?

(O(n) : Orthogonal group, SO(n) : Special orthogonal group)

My attempt for (1) :Let's pick any element C from $\mathrm{O}(n)-\mathrm{S}\mathrm{O}(n)$, and let

$$\begin{array}{rl}& f:\mathrm{S}\mathrm{O}(n)\to \mathrm{S}\mathrm{O}(n);\phantom{\rule{thickmathspace}{0ex}}f(A)=A\\ & g:\mathrm{O}(n)-\mathrm{S}\mathrm{O}(n)\to \mathrm{S}\mathrm{O}(n);\phantom{\rule{thickmathspace}{0ex}}g(A)=CA\end{array}$$

Then f and g are homeomorphisms. Since SO(n) is compact, ${g}^{-1}(\mathrm{S}\mathrm{O}(n))=\mathrm{O}(n)-\mathrm{S}\mathrm{O}(n)$ is also compact. So, they are closed in the subspace topology on O(n). Thus, by the glueing lemma, the function

$$\phi :\mathrm{O}(n)\to \mathrm{S}\mathrm{O}(n)\times {\mathbb{Z}}_{2};\phantom{\rule{thickmathspace}{0ex}}\phi (A)=\{\begin{array}{ll}(f(A),0)\phantom{\rule{1em}{0ex}}& (\text{when}A\in \mathrm{S}\mathrm{O}(n))\\ (g(A),1)& (\text{otherwise})\end{array}$$

becomes continuous. We can easily confirm that $\phi $ is bijective and ${\phi}^{-1}$ is continuous.

Questions:

(1) I'm not sure I'm going in the right direction (in proving (1)). I used glueing lemma to justifying continuousness of the function $\phi $, but I doubt whether I'm using that lemma in the right place.

(2) I tried to prove that two groups are NOT isomorphic, but I can't find any key for that. I tried to compare orders of elements in two groups, but I couldn't find some good elements to compare... Is there any simple way to check whether two groups are isomorphic?

(O(n) : Orthogonal group, SO(n) : Special orthogonal group)

My attempt for (1) :Let's pick any element C from $\mathrm{O}(n)-\mathrm{S}\mathrm{O}(n)$, and let

$$\begin{array}{rl}& f:\mathrm{S}\mathrm{O}(n)\to \mathrm{S}\mathrm{O}(n);\phantom{\rule{thickmathspace}{0ex}}f(A)=A\\ & g:\mathrm{O}(n)-\mathrm{S}\mathrm{O}(n)\to \mathrm{S}\mathrm{O}(n);\phantom{\rule{thickmathspace}{0ex}}g(A)=CA\end{array}$$

Then f and g are homeomorphisms. Since SO(n) is compact, ${g}^{-1}(\mathrm{S}\mathrm{O}(n))=\mathrm{O}(n)-\mathrm{S}\mathrm{O}(n)$ is also compact. So, they are closed in the subspace topology on O(n). Thus, by the glueing lemma, the function

$$\phi :\mathrm{O}(n)\to \mathrm{S}\mathrm{O}(n)\times {\mathbb{Z}}_{2};\phantom{\rule{thickmathspace}{0ex}}\phi (A)=\{\begin{array}{ll}(f(A),0)\phantom{\rule{1em}{0ex}}& (\text{when}A\in \mathrm{S}\mathrm{O}(n))\\ (g(A),1)& (\text{otherwise})\end{array}$$

becomes continuous. We can easily confirm that $\phi $ is bijective and ${\phi}^{-1}$ is continuous.

Questions:

(1) I'm not sure I'm going in the right direction (in proving (1)). I used glueing lemma to justifying continuousness of the function $\phi $, but I doubt whether I'm using that lemma in the right place.

(2) I tried to prove that two groups are NOT isomorphic, but I can't find any key for that. I tried to compare orders of elements in two groups, but I couldn't find some good elements to compare... Is there any simple way to check whether two groups are isomorphic?

College StatisticsAnswered question

clovnerie0q 2022-09-30

How can I show that PGL(2,9) is not isomorphic to ${S}_{6}$?

My primary idea is to compare the size of conjugacy classes of two well-chosen elements in these groups. Is there another simpler approach?

My primary idea is to compare the size of conjugacy classes of two well-chosen elements in these groups. Is there another simpler approach?

College StatisticsAnswered question

misyjny76 2022-09-26

Fundamental group of GL(n,C) is isomorphic to Z. How to learn to prove facts like this?

I know, fundamental group of GL(n,C) is isomorphic to Z. Actually, I've succeed in proving this, but my proof is two pages long and very technical. I want

1. to find some better proofs of this fact (in order to compare to mine);

2. to find some book or article, using which I can learn, how to calculate fundamental groups and, more generally, connectedness components of maps from one space to another;

3. to find something for the reference, which I can use in order to learn, how to write this proofs nicely, using standard terminology.

I know, fundamental group of GL(n,C) is isomorphic to Z. Actually, I've succeed in proving this, but my proof is two pages long and very technical. I want

1. to find some better proofs of this fact (in order to compare to mine);

2. to find some book or article, using which I can learn, how to calculate fundamental groups and, more generally, connectedness components of maps from one space to another;

3. to find something for the reference, which I can use in order to learn, how to write this proofs nicely, using standard terminology.

College StatisticsAnswered question

babuliaam 2022-09-26

How to show that these groups are isomorphic?

Show that group of all real matrices of form

$\left[\begin{array}{cc}x& y\\ -y& x\end{array}\right],\phantom{\rule{2em}{0ex}}(x,y)\ne (0,0)$

is isomorphic with/to C∖{0} under complex multiplication?

I know two ways to show isomorphism: 1) finding a homomorphic function 2) writing the multiplication table and comparing.

Show that group of all real matrices of form

$\left[\begin{array}{cc}x& y\\ -y& x\end{array}\right],\phantom{\rule{2em}{0ex}}(x,y)\ne (0,0)$

is isomorphic with/to C∖{0} under complex multiplication?

I know two ways to show isomorphism: 1) finding a homomorphic function 2) writing the multiplication table and comparing.

College StatisticsAnswered question

Kolby Castillo 2022-09-26

Let C be a chain complex over a commutative ring R. Let M be a subchain of C such that M is chain equivalent to the zero chain complex. Must C/M and C be chain equivalent ?

College StatisticsAnswered question

clasicaacyx 2022-09-25

Is a group isomorphic to one of its realizations?

Suppose I give you an "abstract group" G with the exact same multiplication table as the set $\{\pm 1,\pm i,\pm j,\pm k\}$ (quaternions) under multiplication. And someone says find a homomorphism for this abstract group.

Since I know it follows the same multiplication table as the set I gave above can I suggest the appropriate bijection for those two groups, and say I've constructed the homomorphism?

Suppose I give you an "abstract group" G with the exact same multiplication table as the set $\{\pm 1,\pm i,\pm j,\pm k\}$ (quaternions) under multiplication. And someone says find a homomorphism for this abstract group.

Since I know it follows the same multiplication table as the set I gave above can I suggest the appropriate bijection for those two groups, and say I've constructed the homomorphism?

College StatisticsAnswered question

Alexus Deleon 2022-09-25

Intro to probability chapter 4 ex 31

A group of 50 people are comparing their birthdays (as usual, assume their birthdays are independent, are not February 29, etc.). Find the expected number of pairs of people with the same birthday, and the expected number of days in the year on which at least two of these people were born.

Solution: Creating an indicator r.v. for each pair of people, we have that the expected number of pairs of people with the same birthday is (50C2 . 1/365) by linearity. Now create an indicator r.v. for each day of the year, taking the value 1 if at least two of the people were born that day (and 0 otherwise). Then the expected number of days on which at least two people were born is

$365(1-(364/365{)}^{50}-50\cdot (1/365)\cdot (364/365{)}^{49})$

Could someone explain how we got the answer ?

A group of 50 people are comparing their birthdays (as usual, assume their birthdays are independent, are not February 29, etc.). Find the expected number of pairs of people with the same birthday, and the expected number of days in the year on which at least two of these people were born.

Solution: Creating an indicator r.v. for each pair of people, we have that the expected number of pairs of people with the same birthday is (50C2 . 1/365) by linearity. Now create an indicator r.v. for each day of the year, taking the value 1 if at least two of the people were born that day (and 0 otherwise). Then the expected number of days on which at least two people were born is

$365(1-(364/365{)}^{50}-50\cdot (1/365)\cdot (364/365{)}^{49})$

Could someone explain how we got the answer ?

College StatisticsAnswered question

Jazmyn Pugh 2022-09-24

Is ${\mathbb{Z}}_{84}\oplus {\mathbb{Z}}_{72}$ isomorphic to ${\mathbb{Z}}_{36}\oplus {\mathbb{Z}}_{168}$?

These two groups have the same order. Also, we cannot show that the groups are not isomorphic by comparing the order of groups' elements. Thus, it seems that these two groups are isomorphic.

These two groups have the same order. Also, we cannot show that the groups are not isomorphic by comparing the order of groups' elements. Thus, it seems that these two groups are isomorphic.

College StatisticsAnswered question

clasicaacyx 2022-09-23

Homomorphisms between different types of groups

I've been self-studying some Algebra, and recently did a set of exercises proving both that, for a surjective homomorphism $\varphi $ between two groups G and H, that if:

(i) G is cyclic, then H is cyclic, and;

(ii) G is abelian, then H is abelian.

Proving these two was simple enough, but now I'm thinking about removing the surjectivity constraint on $\varphi $. It does not seem to me that these statements necessarily hold, in either case, but I am having some difficulty coming up with solid counterexamples to either.

I was considering using the Klein group as my H group for part (i), and the dihedral group of order 3 (${D}_{3}$) for part (ii), but I was getting stuck here, as I'm not sure precisely to what I should compare it in order to construct . Any hints/suggestions as to where to look would be greatly appreciated.

I've been self-studying some Algebra, and recently did a set of exercises proving both that, for a surjective homomorphism $\varphi $ between two groups G and H, that if:

(i) G is cyclic, then H is cyclic, and;

(ii) G is abelian, then H is abelian.

Proving these two was simple enough, but now I'm thinking about removing the surjectivity constraint on $\varphi $. It does not seem to me that these statements necessarily hold, in either case, but I am having some difficulty coming up with solid counterexamples to either.

I was considering using the Klein group as my H group for part (i), and the dihedral group of order 3 (${D}_{3}$) for part (ii), but I was getting stuck here, as I'm not sure precisely to what I should compare it in order to construct . Any hints/suggestions as to where to look would be greatly appreciated.

College StatisticsAnswered question

Aubrie Aguilar 2022-09-23

Is the growth on an exponential function faster than on a power function? Is this explanation valid?

I'm trying to find simple mathematical proof that lengthy passwords are more secure than complex (with multiple symbols) ones. I don't want to get into the concept of entropy - it is not useful to me atm. I've found this explanation at Security Stack, which it seemed pretty reasonable to me. My question is: is this explanation valid and, if so, is there a limit to it's validity?. My doubt comes from testing it out considering a password with n digits, but restrict to a group of 26 characters (the alphabet) vs a password with fixed 10 digits, with n possible characters:

$$f(n)={26}^{n}\phantom{\rule{2em}{0ex}}\text{vs}\phantom{\rule{2em}{0ex}}f(n)={n}^{10}.$$

For these two functions it just doesn't work out as the answer in Security Stack explained. So, how does this work?

I guess the title may be a bit misleading, because this question markes as duplicate does not answer my question! I get that exponential growth is faster than polynomial growth, but why doesn't it apply to these two functions I'm comparing? I mean, what is wrong in my line of thought?

I'm trying to find simple mathematical proof that lengthy passwords are more secure than complex (with multiple symbols) ones. I don't want to get into the concept of entropy - it is not useful to me atm. I've found this explanation at Security Stack, which it seemed pretty reasonable to me. My question is: is this explanation valid and, if so, is there a limit to it's validity?. My doubt comes from testing it out considering a password with n digits, but restrict to a group of 26 characters (the alphabet) vs a password with fixed 10 digits, with n possible characters:

$$f(n)={26}^{n}\phantom{\rule{2em}{0ex}}\text{vs}\phantom{\rule{2em}{0ex}}f(n)={n}^{10}.$$

For these two functions it just doesn't work out as the answer in Security Stack explained. So, how does this work?

I guess the title may be a bit misleading, because this question markes as duplicate does not answer my question! I get that exponential growth is faster than polynomial growth, but why doesn't it apply to these two functions I'm comparing? I mean, what is wrong in my line of thought?

College StatisticsAnswered question

Jaqueline Velez 2022-09-22

How large can the outer automorphism group be?

You can make the outer automorphism group very large by taking G be to be a vector space (say over a finite field) so that if $|G|={q}^{n}$, then $Out(G)=G{L}_{n}({\mathbb{F}}_{q})$ of size exponential in n.

So I have two questions related to this:

1) Is there a "family of groups" with a faster growing outer automorphism group?

2) What does the outer automorphism group of a "typical" group look like?

I am leaving both questions fairly vague since I am not sure exactly what kind of an answer I am looking for. The goal is to simply understand how much bigger the outer automorphism group is (compared to the group) in general.

You can make the outer automorphism group very large by taking G be to be a vector space (say over a finite field) so that if $|G|={q}^{n}$, then $Out(G)=G{L}_{n}({\mathbb{F}}_{q})$ of size exponential in n.

So I have two questions related to this:

1) Is there a "family of groups" with a faster growing outer automorphism group?

2) What does the outer automorphism group of a "typical" group look like?

I am leaving both questions fairly vague since I am not sure exactly what kind of an answer I am looking for. The goal is to simply understand how much bigger the outer automorphism group is (compared to the group) in general.

College StatisticsAnswered question

Liberty Page 2022-09-20

Nonisomorph groups of order 2002

While searching for non-isomorph subgroups of order 2002 I just encountered something, which I want to understand. Obviously I looked for abelian subgroups first and found $2002={2}^{2}\ast 503$ so we have the groups

$$\mathbb{Z}/{2}^{2}\mathbb{Z}\times \mathbb{Z}/503\mathbb{Z},\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/503\mathbb{Z}$$

Now I want to understand why those two are not isomorph. I know that for two groups $\mathbb{Z}/n\mathbb{Z}\times \mathbb{Z}/m\mathbb{Z}\cong \mathbb{Z}/(nm)\mathbb{Z}$ it has to hold that $gcd(n,m)=1$. But I don't understand how we can compare Groups written as two products with groups written as three products as above, how does that work? And I think that goes in the same direction: How is it then at the same time that

$$\mathbb{Z}/4\mathbb{Z}\times \mathbb{Z}/503\mathbb{Z}\cong \mathbb{Z}/2012\mathbb{Z}\u2246\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/503\mathbb{Z}\cong \mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/1006Z$$

because $gcd(4,2012)\ne 1,gcd(2,2)\ne 1,gcd(503,1006)\ne 1$. I don't understand the difference to the first comparison.

While searching for non-isomorph subgroups of order 2002 I just encountered something, which I want to understand. Obviously I looked for abelian subgroups first and found $2002={2}^{2}\ast 503$ so we have the groups

$$\mathbb{Z}/{2}^{2}\mathbb{Z}\times \mathbb{Z}/503\mathbb{Z},\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/503\mathbb{Z}$$

Now I want to understand why those two are not isomorph. I know that for two groups $\mathbb{Z}/n\mathbb{Z}\times \mathbb{Z}/m\mathbb{Z}\cong \mathbb{Z}/(nm)\mathbb{Z}$ it has to hold that $gcd(n,m)=1$. But I don't understand how we can compare Groups written as two products with groups written as three products as above, how does that work? And I think that goes in the same direction: How is it then at the same time that

$$\mathbb{Z}/4\mathbb{Z}\times \mathbb{Z}/503\mathbb{Z}\cong \mathbb{Z}/2012\mathbb{Z}\u2246\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/503\mathbb{Z}\cong \mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/1006Z$$

because $gcd(4,2012)\ne 1,gcd(2,2)\ne 1,gcd(503,1006)\ne 1$. I don't understand the difference to the first comparison.

College StatisticsAnswered question

skauvzc 2022-09-17

Difference of fixed points by subgroup action

If G is a finite group and X is a finite G-set, then we have a class equation which tells us

$$|X|=|{X}^{G}|+\sum |G/{G}_{x}|$$

where ${X}^{G}$ is the set of G-fixed points, ${G}_{x}$ is the stabilizer of $x\in X$ and the sum is taken over representative elements of classes of non-singleton G-orbits.

My question, which arise from a proof in a paper which seems to use precisely this fact, is the following. If $H\le G$ is a normal subgroup, are we able to write the cartinality of ${X}^{H}\setminus {X}^{G}$ as sum of cardinalities of G/H orbits? If this is the case, over what that sum is taken?

I guess the result should descend by comparing the two class equations for H and G, but I'm struggling to write that down.

If G is a finite group and X is a finite G-set, then we have a class equation which tells us

$$|X|=|{X}^{G}|+\sum |G/{G}_{x}|$$

where ${X}^{G}$ is the set of G-fixed points, ${G}_{x}$ is the stabilizer of $x\in X$ and the sum is taken over representative elements of classes of non-singleton G-orbits.

My question, which arise from a proof in a paper which seems to use precisely this fact, is the following. If $H\le G$ is a normal subgroup, are we able to write the cartinality of ${X}^{H}\setminus {X}^{G}$ as sum of cardinalities of G/H orbits? If this is the case, over what that sum is taken?

I guess the result should descend by comparing the two class equations for H and G, but I'm struggling to write that down.

College StatisticsAnswered question

driliwra7 2022-09-14

When is examining only one direction of a bijective map $\varphi :X\to Y$ enough to categorize it as an isomorphism?

Let X,Y be sets with $\varphi :X\to Y$ being bijective. If we consider X and Y various structures and ask what conditions do we have to impose on $\varphi $ for it to be an isomorphism, we can break these cases up into two groups:

Group 1:

- If X,Y are vector spaces, then $\varphi $ needs to be linear.

- If X,Y are groups, then $\varphi $ needs to respect the group operation.

- If X,Y are metric spaces, then $\varphi $ needs to be an isometry.

- etc.

Group 2:

- If X,Y are topological spaces, then $\varphi $ needs to be continuous and open.- If X,Y are smooth manifolds, then $\varphi $ needs to be a diffeomorphism.

- etc.

In case one, we only need to verify that $\varphi $ satisfies certain properties (that $\varphi $ is linear, respects group operation, isometry, etc.). Whereas in group 2, we need to verify that both $\varphi $ AND ${\varphi}^{-1}$ satisfy certain properties (continuous, smooth, etc). My question is, what is it that distinguishes the members of group 1 verses group 2? Why is it enough to check a certain condition in one direction only enough in certain cases?

Let X,Y be sets with $\varphi :X\to Y$ being bijective. If we consider X and Y various structures and ask what conditions do we have to impose on $\varphi $ for it to be an isomorphism, we can break these cases up into two groups:

Group 1:

- If X,Y are vector spaces, then $\varphi $ needs to be linear.

- If X,Y are groups, then $\varphi $ needs to respect the group operation.

- If X,Y are metric spaces, then $\varphi $ needs to be an isometry.

- etc.

Group 2:

- If X,Y are topological spaces, then $\varphi $ needs to be continuous and open.- If X,Y are smooth manifolds, then $\varphi $ needs to be a diffeomorphism.

- etc.

In case one, we only need to verify that $\varphi $ satisfies certain properties (that $\varphi $ is linear, respects group operation, isometry, etc.). Whereas in group 2, we need to verify that both $\varphi $ AND ${\varphi}^{-1}$ satisfy certain properties (continuous, smooth, etc). My question is, what is it that distinguishes the members of group 1 verses group 2? Why is it enough to check a certain condition in one direction only enough in certain cases?

Comparing proportions between two groups examples is one of those popular tasks that students have to work with when they have two samples of statistical data that must be compared. As a rule, the questions like those you can see below are based on college-level statistical analysis, yet some tasks will require more. See our helpful answers and see what formulas and equations can be used to cut the corners as you achieve a suitable solution. These are only complex when you do have statistics examples that will guide you further.