1

I need a proof of the following fact: if a group $G=M \times N$ is the direct product of subgroups $M$ and $N$ such that $\lvert M \rvert$ and $\lvert N \rvert$ are relatively prime, then $\mathrm{Aut}(G) = \mathrm{Aut}(M) \times \mathrm{Aut}(N)$.

This is the way the result is recalled on some lecture notes I am reading, I think $M$ and $N$ are implicitly supposed finite (maybe $G$ also?) otherwise this does not make sense.

This should be rather standard (it is used but not proven here), but I've taken a quick look on Rotman, Roman and Robinson but none of them seems to prove this. Do you have a reference for the proof/can you share here a proof you know?

Thanks

carciofo21
  • 569
  • 3
  • 13
  • Hint: if $g$ is an automorphism on $G$, what can you say about $g(x, 1)$, and $g(1, y)$. – Rob Arthan Nov 30 '20 at 22:43
  • Well, call $\phi$ an automorphism of $G$, then for every $m \in M$ the order of $\phi(m)$ must divide $o(m)$, which divides $\lvert M \rvert$. So either $m=1$ or $\phi(m) \notin N$. Hence every automorphism of $G$ restricts to an automorphism of $M$ and the same holds for $N$. The map $\phi \mapsto (\phi|_M, \phi|_N)$ is well-posed and readily checked to be an homomorphism, whose inverse is easily constructed via the universal property of the coproduct. (Or directly with association $(\phi_M, \phi_N) \mapsto \phi$ where $\phi(g)=\phi(mn)=\phi_M(m)\phi_N(n)$ for every $g=mn \in G$). – carciofo21 Nov 30 '20 at 23:46
  • Anyway I was asking for a reference, not for an hint. – carciofo21 Nov 30 '20 at 23:47
  • You also asked for a proof: I have given you a hint towards a proof. – Rob Arthan Nov 30 '20 at 23:49
  • Here’s a reference. The linked-to duplicate also has a proof. – Arturo Magidin Dec 01 '20 at 01:32
  • @ArturoMagidin Thank you very much for the reference. Anyway the "duplicate" - which I was aware of and which I cited in my very question - does not contain a proof of the result I was looking for but just the statement "Okay, so I know that if $G,H$ are two groups whose orders are relatively prime, then $\mathrm{Aut}(G×H)\simeq \mathrm{Aut}(G)×\mathrm{Aut}(H).$" Then it goes on on more specific matters without recalling the proof pf the result mentioned. – carciofo21 Dec 01 '20 at 09:52
  • The duplicate mentions the paper I linked to (and its second part). – Arturo Magidin Dec 01 '20 at 11:36

1 Answers1

0

Let $n=|M|, m=|M|$ there exists integers $a,b$ such that $am+bn=1$. Let $x\in M$, write $f(x,1)=(u,v)$. We have $f((x,1)^{am})=f(x^{am},1)=f(1,1)=(1,1)=f(u^{am},v^{am})$.

We have $u^{am}=(u^m)^a=1$, $v^{am}=v^{1-bn}=vv^{-bn}=v$, we deduce that $(1,1)=(1,v), v=1$. Similarly, we show that for every $y\in N$, $f(1,y)=(1,l)$. This implies that $f(x,y)=(g(x),h(y))$ and $f^{-1}(x,y)=(g'(x),h'(y))$. $g'$ is the inverse of $g$ and $h'$ the inverse of $h$.

Tsemo Aristide
  • 87,475