# Let $G$ to be abelian group and $|G|=mn$ when $(m,n)=1$. $G_m=\{g\mid g^m=e\}$,$G_n=\{g\mid g^n=e\}$, prove isomorphism

I want to prove $f:G_n\times G_m\rightarrow G$ when $f(g,h)=gh$ is an isomorphism.

First of all I showed that $G_m,G_n$ are subgroups of $G$ (easy).

Now I want to show that for every $a,b, \in G$, $f(ab)=f(a)f(b)$.

Let $a=(g_1,h_1)$ and $b=(g_2,h_2)$

$\implies f(g_1,h_1)=g_1h_1$ , $f(g_2,h_2)=g_2h_2$

$\therefore f(g_1,h_1) f(g_2,h_2)=g_1h_1g_2h_2=g_1g_2h_1h_2$ (because G is abelian) $=f\bigl((g_1g_2),(h_1h_2)\bigr)$

Then, I need to show that only $f(e)=e$.

Because $(m,n)=1$, only $f(e,e) = e e = e$.

Am I right? If not how can I prove this?

Is $f$ an isomorphism even if $G$ is not abelian?

• Very nice posed question, just need a little better formating +1 Jan 1, 2015 at 18:14
• You probably want to change the word "automorphism" for "isomorphism", unless you mean the internal direct product with $\times$, but then it becomes a bit too much for no reason. Normally you use the word automorphism when you can write $f : G \to G$. Jan 1, 2015 at 20:55

Suppose

$$f(a,b):=ab=e\implies a=b^{-1}$$

Buth te last equality is impossible as $\;a\in G_n\;,\;\;b\in G_m\;$ and thus the only possible element in both of them is the unity, i.e. $\;G_n\cap G_m=\{e\}\;$ .

For a counter example with $\;G\;$ non abelian take $\;G=S_3\;$ , though in this case $\;G_2\;$ is not a subgroup.

• This shows that $f$ is injective, and the OP has shown that it is a homomorphism. But what about surjectivity?
– user169852
Jan 1, 2015 at 21:18
• @Bungo $\;|G|=mn=|G_m||G_n|\implies$ a map between these two is injective iff it is surjective iff it is bijective. Jan 1, 2015 at 22:07
• Sure, but is it immediately obvious that $|G_m||G_n| = mn$? I don't see where the OP has proven this (or even stated it) so I just wanted to point out that it requires proof in case it was overlooked.
– user169852
Jan 1, 2015 at 22:18
• @Bungo I think we shall we leave some work for the OP to complete, shalln't we? After all, it may be not obvious but it is very, very easy to prove. Jan 2, 2015 at 0:28
• @Timbuc it was easy to follow you answers, thanks. but its stil unclear to me how the counter example works. $s3 = \{ e,a,b,a^2,ab,a^2b\}$ (my knowledge on S3 is minimal so if you could expand your answer i would be thankful Jan 2, 2015 at 8:59