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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
$\mathbb{Z}/{2}^{2}\mathbb{Z}×\mathbb{Z}/503\mathbb{Z},\mathbb{Z}/2\mathbb{Z}×\mathbb{Z}/2\mathbb{Z}×\mathbb{Z}/503\mathbb{Z}$
Now I want to understand why those two are not isomorph. I know that for two groups $\mathbb{Z}/n\mathbb{Z}×\mathbb{Z}/m\mathbb{Z}\cong \mathbb{Z}/\left(nm\right)\mathbb{Z}$ it has to hold that $gcd\left(n,m\right)=1$. 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
$\mathbb{Z}/4\mathbb{Z}×\mathbb{Z}/503\mathbb{Z}\cong \mathbb{Z}/2012\mathbb{Z}≆\mathbb{Z}/2\mathbb{Z}×\mathbb{Z}/2\mathbb{Z}×\mathbb{Z}/503\mathbb{Z}\cong \mathbb{Z}/2\mathbb{Z}×\mathbb{Z}/1006Z$
because $gcd\left(4,2012\right)\ne 1,gcd\left(2,2\right)\ne 1,gcd\left(503,1006\right)\ne 1$. I don't understand the difference to the first comparison.

### Answer & Explanation

espovilham7

Beginner2022-09-21Added 10 answers

Step 1
Here's where the hypothesis $gcd\left(n,m\right)=1$ plays a role : if $d:=gcd\left(n,m\right)\ne 1$, then $\mathbb{Z}/nm\mathbb{Z}$ has an element of order ${d}^{2}$ but $\mathbb{Z}/n\mathbb{Z}×\mathbb{Z}/m\mathbb{Z}$ does not.
Step 2
In your example, $\mathbb{Z}/4\mathbb{Z}×\mathbb{Z}/503\mathbb{Z}$ has an element of order 4 (namely (1,0)), but $\mathbb{Z}/2\mathbb{Z}×\mathbb{Z}/1006\mathbb{Z}$ 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
${\mathbb{Z}}_{2002}\cong {\mathbb{Z}}_{2}×{\mathbb{Z}}_{7}×{\mathbb{Z}}_{11}×{\mathbb{Z}}_{13}$
Abelian groups of order 2012:
Since $2012={2}^{2}\cdot 503$ it might initially seem like there are four possibilities:
${\mathbb{Z}}_{2}×{\mathbb{Z}}_{2}×{\mathbb{Z}}_{503}$
${\mathbb{Z}}_{4}×{\mathbb{Z}}_{503}$
${\mathbb{Z}}_{2}×{\mathbb{Z}}_{1006}$
${\mathbb{Z}}_{2012}$
Step 2
But using the fact that
${\mathbb{Z}}_{n}×{\mathbb{Z}}_{m}\cong {\mathbb{Z}}_{nm}\phantom{\rule{1em}{0ex}}\text{iff}\phantom{\rule{1em}{0ex}}GCD\left(n,m\right)=1.$
we get that
${\mathbb{Z}}_{2}×{\mathbb{Z}}_{2}×{\mathbb{Z}}_{503}\cong {\mathbb{Z}}_{2}×{\mathbb{Z}}_{1006}$
and
${\mathbb{Z}}_{4}×{\mathbb{Z}}_{503}\cong {\mathbb{Z}}_{2012}$
and
${\mathbb{Z}}_{2}×{\mathbb{Z}}_{1006}≆{\mathbb{Z}}_{2012}$
Hence there are exactly two non-isomorphic groups of order 2012.

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