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Let $A$ be an abelian finite group and $f$ such automorphism that $f^2(a)=a^{-1}$ for all $a \in A$.

Describe all pairs $(A, f)$ up to isomorphism.

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    The notation $f^2(a)$ means $$f(f(a))$$ $$f(a)f(a)$$ Which? – 2017-02-05
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    $f^2(a)=f(f(a))$ – 2017-02-05
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    To get started, assume $A$ is cyclic, isomorphic to $\mathbb{Z}_n$, and suppose $f(1) = m$, What can you deduce about the relationship between $m$ and $n$? Is it if and only if? Next, suppose A is a direct sum of cyclic groups. – 2017-02-05
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    No. The condition $(m,n) = 1$ is necessary and sufficient for $f$ to be an automorphism, but not enough to force $f^2(a)=a$. You haven't used the hypothesis. – 2017-02-05
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    We can say that $(m,n)=1$. Yes, if and only if. Because $Aut(Z_n)=(Z_n)^*$ If $A$ is a direct sum of cyclic, i.e. $A=Z_{n_1} \oplus ... \oplus Z_{n_k}$, then $f(1, 1, .., 1)=(b_1, b_2, .., b_k)$ and $(b_i, n_i)=1$ – 2017-02-05
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    Suppose $A=Z_n$, then $a \to f(a) \to n-a$ and $(a, n)=1$, $(n-a, n)=1$ hence $n$ is a prime number. Is it right? – 2017-02-05
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    No, it's not right. Assume $f(1) = m$, where $(m,n)=1$. Now assume $f^2(a) = -a$, for all $a \in A$. Expand that equation. You get a relation between $m$ and $n$, – 2017-02-05
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    By the way, a few comments ago, I wrote $f^2(a) = a$, but I that was a typo. I meant $f^2(a) = -a$. – 2017-02-05
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    I made a mistake, I mean $a \to b \to n-a$ and $(b, n)=(n-a, n)=1$ For the your equation: $1 \to m \to n-1$ and $(m, n)=(n-1, n)=1$ – 2017-02-05
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    If you don't see it immediately, try a few small values of $n$ and actually _find_ the automorphisms $f$ (i.e., find the possible the values of m = f(1)) of $\mathbb{Z}_n$ such that $f^2(a)=-a$. – 2017-02-05
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    I have to leave for now -- hope I helped a little. I'll be back later tonight; – 2017-02-05
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    If $A=Z_n$, then there's $a \in (Z_n)^*$ such that $ord(a)=4$ and hence $n \equiv 1 \mod 4$. Is it right? – 2017-02-06
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    No. If $a \in (Z_n)^*$ is such that $\text{ord}(a)=4$, that implies $a|\phi(n)$. It doesn't imply $n \equiv 1 \pmod 4$, – 2017-02-06
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    My suggestion was to take some small values of $n$, say $n = 2,3,4,5,6$, and for each of those values of $n$, find all automorphisms $f$ of $\mathbb{Z}_n$ such that $f^2(a) = -a$. – 2017-02-06
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    Can I ask where the problem comes from? – 2017-02-06
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    In my comment where I wrote "that implies $a|\phi(n)$", I meant to write "that implies $4|\phi(n)$". – 2017-02-06
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    For $Z_2$ it's $f=id$, for $Z_5$ it's $f(1)=2$ or $f(1)=3$. Other groups have no such automorphisms. I meant $n \equiv 1 \mod 4$ if $n$ is prime. This problem I found in my university. – 2017-02-06
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    If $A$ is a finite abelian group isomorphic to $\mathbb{Z}_n$ where n is an odd prime, then yes, there exists an automorphism $f$ of $A$ such that $f^2(a) = -a$ for all $a \in A$ if and only if $n \equiv 1 \pmod 4$. – 2017-02-06

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