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$G_1$ and $G_2$ are finite groups and $K \leq G_1 \times G_2$.

$H_1 = \left\{ g \in G_1 : (g,e) \in K \right\}$

$H_2 = \left\{ g \in G_2 : (e,g) \in K \right\}$

I've already shown that $H_1 \leq G_1$, $H_2 \leq G_2$ and $H_1 \times H_2 \leq K$.

Now suppose that $|G_1|$ and $|G_2|$ are coprime. Show that $K = H_1 \times H_2$ and finally show that this result need not follow if $|G_1|$ and $|G_2|$ are not coprime.

As $|G_1|$ and $|G_2|$ are coprime we must have that $|H_1| \neq |H_2|$ as by Lagrange's theorem $|H_1| \ \mid \ |G_1|$ and $|H_2| \ \mid \ |G_2|$

Also $|G_1|$ and $|G_2|$ are coprime implies that $|H_1|$ and $|H_2|$ are coprime which, in turn, implies that $| H_1 \times H_2 | = |H_1| \cdot |H_2|$

As $H_1 \times H_2 \leq K$ we must have, by Lagrange's theorem, that $| H_1 \times H_2 | = |H_1| \cdot |H_2| \ \mid \ K$

Also as $K \leq G_1 \times G_2$ and $|G_1 \times G_2| = |G_1| \cdot |G_2|$ then $K \ \mid \ |G_1| \cdot |G_2|$

I can't see the connection to make to conclude that $K = H_1 \times H_2$, I'm pretty sure that if $|K| = |H_1 \times H_2| = |H_1| \cdot |H_2|$ it implies the result, but I can't see how to conclude it. Also, in regards to the second part, showing that the result need not follow if $|G_1|$ and $|G_2|$ are not coprime, this is obvious, but is there a 'simple' counter example or can it be argued in a similar fashion to the above?

Thanks!

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2 Answers 2

up vote 5 down vote accepted

Let $(g_1,g_2)\in K$ be arbitrary. You want to prove that $g_1\in H_1$ and $g_2\in H_2$.

Hints:

  1. Let $m$ be the order of $g_2$. What can you say about $(g_1,g_2)^m$?
  2. Show that $g_1$ and $g_1^m$ generate the same subgroup of $G_1$.
  3. Conclude that $g_1\in H_1$.

For the counter example when co-primality assumption is dropped consider $G_1=G_2=\mathbb{Z}/4\mathbb{Z}$ and $K$ the (additive) subgroup generated by $(\overline{2},\overline{2})$.

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If $m$ is the order of $g_2$ then $(g_1,g_2)^m = (g_1^m, e)$ so $g_1^m \in H_1$, If the order of $g_1$ is $p$ then letting $x = lcm(m, p+1)$ we get that, since $K$ is closed under multiplication, $(g_1,g_2)^x = (g_1, e)$ and so $g_1 \in H_1$. We can do the same for $g_2$ and conclude that $K \leq H_1 \times H_2$ and so $K = H_1 \times H_2$. Is that correct? Thanks! –  Noble. Mar 18 '13 at 17:01
    
You are getting there, but not quite. Simply because $x$ is a multiple of $p+1$ you won't get that $g_1^x=g_1.$ To that end you need $x\equiv 1\pmod{p}$. So you should use Bezout's identity instead. There exist integers $u,v$ such that $up+vm=1$. Then $$(g_1,g_2)^{vm}=(g_1^{vm},g_2^{vm})=(g_1^{1-up},e)=(g_1,e).$$ –  Jyrki Lahtonen Mar 18 '13 at 17:38
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Sorry, yes, that was rather silly of me to think if $x$ is a multiple of $p+1$ then $g_1^x = g_1$. Thanks a lot! –  Noble. Mar 18 '13 at 17:44
    
Sorry again, I know Bezout's identity but I thought if $x = 1 (\mod p)$ then there exist integers $u,v$ such that $ux + vp = 1$, how do we jump to $up + vm = 1$, is it because we want $x$ to be a multiple of $p$? It looks like the $hcf(m,p) = 1$ but I don't see how this follows. Thanks once again. –  Noble. Mar 18 '13 at 17:53
    
Because $\gcd(p,m)=1$ Bezout gives us the existence of $u$ and $v$ such that $up+vm=1$. Then you can use $x=vm$, because $x$ will 1) be a multiple of $m$, and 2) $x\equiv1\pmod{p}$. –  Jyrki Lahtonen Mar 18 '13 at 18:00

We can always find a counterexample if $|G_1|$ and $|G_2|$ are not coprime.

Let $p$ be a prime dividing $|G_1|$ and $|G_2|$. By Cauchy's theorem $G_1$ and $G_2$ both contain an element of order $p$, let $x_1 \in G_1$ and $x_2 \in G_2$ have order $p$. Then the subgroup generated by $(x_1, x_2)$ is not of the form $A \times B$, where $A \leq G_1$ and $B \leq G_2$.

(The example given by Jyrki is exactly this. The smallest example is $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$)

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Gah! How could I miss the smallest example! +1 –  Jyrki Lahtonen Mar 18 '13 at 16:29

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