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Let $A \ne \{e\}$ be abelian group which is not isomorphic to $Z_2 $. Prove that $Inn(A) \ne Aut(A) $.

first, I have proved that $Inn(A) = \{id_a\} $.

So, we need to prove that A got a not trivial automorphism.

I consider 2 cases:

1) there exist $x \in A $ such that $x \ ^ 2 \ne e $ , in this case I defined $f : A \to A $ $f(a) = a $ for $a\in A $ $a\ne x, x\ ^2 $ , and $f(x) = x \ ^ 2 , f(x \ ^ 2) =x $ , f is indeed automorphism.

2) for each $a \in A $ it is true that $a \ ^ 2 = e $. im not sure how to continue from here

Thanks for helping.

1 Answers1

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Unless $a^2=e$ for all, the map $ x \mapsto x^{-1}$ is a nontrivial automorphism.

If $a^2=e$ for all, then $A$ is a vector space over $\mathbb F_2$ and so has a basis. The map that swaps two elements of that basis is a nontrivial automorphism.

lhf
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  • This argument requires the existence of a basis, which equivalent to the axiom of choice. There may be an argument that avoids that. I think it's unlikely though because one has to deal very large abelian groups, which may not even be finitely generated... – lhf Dec 02 '16 at 23:41
  • @lhs You don't need a full basis, just two nonzero elements. – Michael Burr Dec 02 '16 at 23:47
  • Shouldn't it be, "unless $a^2=e$ for all $a\in A$"? – Camilo Arosemena-Serrato Dec 03 '16 at 00:17
  • @CamiloArosemena, right, fixed. Thanks. – lhf Dec 03 '16 at 00:20
  • Thanks. this is exactly what i thought to do, wasn't sure if im right about the existence of a basis .. @lhf –  Dec 03 '16 at 07:03