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.