How to Prove Aut(H ⊕ K) ≅ Aut(H) ⊕ Aut(K)?

[arthurhenry]

Wikipedia states:
when G splits as direct sum of H and K, then

Aut(H \oplus K) \cong Aut(H) \oplus Aut(K)

Could someone please help me prove this or perhaps give a reference.
Thank you

 

[micromass]

Hi arthurhenry!

Where exactly did you see this, what was the context. In general this is false:

Aut(\mathbb{Z}_p^2)=GL_2(\mathbb{Z}_p)

but

Aut(\mathbb{Z}_p)\times Aut(\mathbb{Z}_p)=\mathbb{Z}_{p-1}^2

 

[arthurhenry]

Hello,


http://en.wikipedia.org/wiki/Abelian_group


here is the link, thank you for the quick response

 

[micromass]

OK, next time, could you please state these things completely?? The wikipedia article says that Aut(H\times K)\cong Aut(H)\times Aut(K) if the groups are finite, abelian and of coprime order! You need those conditions.

As for the proof, try to prove it in these steps

• Given an automorphism f<img src="/styles/physicsforums/xenforo/smilies/arghh.png" class="smilie" loading="lazy" alt=":H" title="Gah! :H" data-shortname=":H" />\times K\rightarrow H\times K, then f(H)=H and f(K)=K.
• We have a homomorphism
  
  \Phi:Aut(H\times K)\rightarrow Aut(H)\times Aut(K):f\rightarrow (f\vert_H, f\vert_K)
  
• Find an inverse of the homomorphism.

 

[arthurhenry]

I am sorry, I realized right after I sent the email; and thank you, now I will work on it.

 

[arthurhenry]

This might be bad, but I have had problem finding an inverse. I am afraid I am not suing all of the hypothesis either.