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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.

espovilham7

Beginner2022-09-21Added 10 answers

Step 1

Here's where the hypothesis $gcd(n,m)=1$ plays a role : if $d:=gcd(n,m)\ne 1$, then $\mathbb{Z}/nm\mathbb{Z}$ has an element of order ${d}^{2}$ but $\mathbb{Z}/n\mathbb{Z}\times \mathbb{Z}/m\mathbb{Z}$ does not.

Step 2

In your example, $\mathbb{Z}/4\mathbb{Z}\times \mathbb{Z}/503\mathbb{Z}$ has an element of order 4 (namely (1,0)), but $\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/1006\mathbb{Z}$ has no element of order 4.

Here's where the hypothesis $gcd(n,m)=1$ plays a role : if $d:=gcd(n,m)\ne 1$, then $\mathbb{Z}/nm\mathbb{Z}$ has an element of order ${d}^{2}$ but $\mathbb{Z}/n\mathbb{Z}\times \mathbb{Z}/m\mathbb{Z}$ does not.

Step 2

In your example, $\mathbb{Z}/4\mathbb{Z}\times \mathbb{Z}/503\mathbb{Z}$ has an element of order 4 (namely (1,0)), but $\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/1006\mathbb{Z}$ has no element of order 4.

Celinamg8

Beginner2022-09-22Added 1 answers

Step 1

First let's note that ${2}^{2}\cdot 503=2012\ne 2002$

Abelian groups of order 2002:

There is only one Abelian group of order 2002, namely

$${\mathbb{Z}}_{2002}\cong {\mathbb{Z}}_{2}\times {\mathbb{Z}}_{7}\times {\mathbb{Z}}_{11}\times {\mathbb{Z}}_{13}$$

Abelian groups of order 2012:

Since $2012={2}^{2}\cdot 503$ it might initially seem like there are four possibilities:

$${\mathbb{Z}}_{2}\times {\mathbb{Z}}_{2}\times {\mathbb{Z}}_{503}$$

$${\mathbb{Z}}_{4}\times {\mathbb{Z}}_{503}$$

$${\mathbb{Z}}_{2}\times {\mathbb{Z}}_{1006}$$

$${\mathbb{Z}}_{2012}$$

Step 2

But using the fact that

$${\mathbb{Z}}_{n}\times {\mathbb{Z}}_{m}\cong {\mathbb{Z}}_{nm}\phantom{\rule{1em}{0ex}}\text{iff}\phantom{\rule{1em}{0ex}}GCD(n,m)=1.$$

we get that

$${\mathbb{Z}}_{2}\times {\mathbb{Z}}_{2}\times {\mathbb{Z}}_{503}\cong {\mathbb{Z}}_{2}\times {\mathbb{Z}}_{1006}$$

and

$${\mathbb{Z}}_{4}\times {\mathbb{Z}}_{503}\cong {\mathbb{Z}}_{2012}$$

and

$${\mathbb{Z}}_{2}\times {\mathbb{Z}}_{1006}\u2246{\mathbb{Z}}_{2012}$$

Hence there are exactly two non-isomorphic groups of order 2012.

First let's note that ${2}^{2}\cdot 503=2012\ne 2002$

Abelian groups of order 2002:

There is only one Abelian group of order 2002, namely

$${\mathbb{Z}}_{2002}\cong {\mathbb{Z}}_{2}\times {\mathbb{Z}}_{7}\times {\mathbb{Z}}_{11}\times {\mathbb{Z}}_{13}$$

Abelian groups of order 2012:

Since $2012={2}^{2}\cdot 503$ it might initially seem like there are four possibilities:

$${\mathbb{Z}}_{2}\times {\mathbb{Z}}_{2}\times {\mathbb{Z}}_{503}$$

$${\mathbb{Z}}_{4}\times {\mathbb{Z}}_{503}$$

$${\mathbb{Z}}_{2}\times {\mathbb{Z}}_{1006}$$

$${\mathbb{Z}}_{2012}$$

Step 2

But using the fact that

$${\mathbb{Z}}_{n}\times {\mathbb{Z}}_{m}\cong {\mathbb{Z}}_{nm}\phantom{\rule{1em}{0ex}}\text{iff}\phantom{\rule{1em}{0ex}}GCD(n,m)=1.$$

we get that

$${\mathbb{Z}}_{2}\times {\mathbb{Z}}_{2}\times {\mathbb{Z}}_{503}\cong {\mathbb{Z}}_{2}\times {\mathbb{Z}}_{1006}$$

and

$${\mathbb{Z}}_{4}\times {\mathbb{Z}}_{503}\cong {\mathbb{Z}}_{2012}$$

and

$${\mathbb{Z}}_{2}\times {\mathbb{Z}}_{1006}\u2246{\mathbb{Z}}_{2012}$$

Hence there are exactly two non-isomorphic groups of order 2012.

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