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?

  • $\begingroup$ Very nice posed question, just need a little better formating +1 $\endgroup$
    – Timbuc
    Jan 1, 2015 at 18:14
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    $\begingroup$ 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 $f : G \to G$. $\endgroup$ Jan 1, 2015 at 20:55

1 Answer 1



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

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.

  • $\begingroup$ This shows that $f$ is injective, and the OP has shown that it is a homomorphism. But what about surjectivity? $\endgroup$
    – user169852
    Jan 1, 2015 at 21:18
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    $\begingroup$ @Bungo $\;|G|=mn=|G_m||G_n|\implies $ a map between these two is injective iff it is surjective iff it is bijective. $\endgroup$
    – Timbuc
    Jan 1, 2015 at 22:07
  • $\begingroup$ 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. $\endgroup$
    – user169852
    Jan 1, 2015 at 22:18
  • $\begingroup$ @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. $\endgroup$
    – Timbuc
    Jan 2, 2015 at 0:28
  • $\begingroup$ @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 $\endgroup$ Jan 2, 2015 at 8:59

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