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Problem: Suppose $G$ is a finite abelian group and there is no non-trivial isomorphism $f: G \rightarrow G$ which has order $2$. Prove that $G$ is either the trivial group $\{1\}$ or isomorphic to $\mathbb Z / 2 \mathbb Z$.

My idea:

Let $f: G \rightarrow G, x \mapsto x^{-1}$, we have

$$f(xy)=(xy)^{-1}=y^{-1}x^{-1}=x^{-1}y^{-1}=f(x)f(y), \forall x,y \in G.$$

Also, $\ker f=\{x \in G: x^{-1}=1\}=1.$ For all $x\in G$, there exists $x^{-1} \in G$ s.t $(x^{-1})^{-1}=x$. Thus, $f$ is isomorphism.

$f$ has order $2$ since $f(f(x))=x, \forall x \in G$, in other words, $f\circ f=\rm id$.

Therefore, $f$ must be trivial, $f=\rm id$, it induces $x^2=1, \forall x\in G$.

Assume that $X\ne \{1\}$, I want to prove that $|X|=2$, but I have no idea.

Help me, thanks!

cqfd
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Thu Le
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    I don't have a complete answer, but note that, e.g. the Klein four-group ($Z_2 \times Z_2$) has 4 elements all of which are self-inverse, so you're not guaranteed that $x \mapsto x^{-1}$ is a nontrivial automorphism on larger groups. – Gregory J. Puleo Jul 12 '19 at 04:15

1 Answers1

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I'll write $G$ additively, rather than multiplicatively. As you say, the automorphism $x\mapsto -x$ has order $2$ unless $G$ is $2$-torsion, that is $x+x=0$ for all $x\in G$. A $2$-torsion Abelian group is effectively a vector space over the field $\Bbb F_2$, so has the form $$G=\Bbb F_2^k=\{(a_1,\ldots,a_k):a_1,\ldots,a_k\in\Bbb F_2\}.$$ When $k\ge2$, $G$ has the non-trivial automorphism $$\phi:(a_1,\ldots,a_k)\mapsto(a_1+a_2,a_2,\ldots,a_k)$$ which is its own inverse. So unless $k\le 1$, $G$ has an automorphism of order $2$.

Angina Seng
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