$\sigma(X) = X$ or $-X$, then $G$ is solvable? Let $G$ be a group and $\sigma(X)$ be a non-trivial automorphism of the group $G$ such that $\sigma(X) = X$ or $X^{-1}$, then prove that $G$ is solvable?
My approach: I calculated that 
$H = \{ X \mid \sigma(X) = X\; \}$ and $K = \{ X \mid \sigma(X) = X^{-1}\; \}$ both are normal subgroup because of the following relation 
$$xyx^{-1} = y^{-1}$$ 
and $K$ is an abelian subgroup. I don't know how to proceed further. Any help would be appreciated. There is no information on the order of the group.
 A: $H$ is clearly a subgroup of $G$. If $H=G$, then $\sigma$ is trivial, contrary to assumption, so $H<G$. 
Let $h\in H$ and $g\in G\setminus H$ so $gh\in G\setminus H$ and $\sigma(gh)=(gh)^{-1}=h^{-1}g^{-1}$. On the other other hand, $\sigma(gh)=\sigma(g)\sigma(h)=g^{-1}h$, so $h^{-1}g^{-1}=g^{-1}h$ or $g^{-1}hg=h^{-1}$. In other words, $g$ acts by inversion on $H$. It follows that $H$ is normal and abelian. (Since a group with inversion as an automorphism must be abelian.) 
If $h^2=1$ for every $h \in H$, then $\sigma(g)=g^{-1}$ for every $g\in G$ and $G$ is abelian, for the same reason as above. We may thus assume that there is $h\in H$ with $h^2\neq 1$. Now, let $g_1,g_2\in G\setminus H$. By the above, $g_1$ and $g_2$ both invert $h$ (by conjugation), so $g_1g_2$ must centralise it. Since $h^2\neq 1$, $g_1g_2$ does not invert $h$ and thus $g_1g_2\notin G\setminus H$  and thus $g_1g_2\in H$. We have shown that, for every two elements not in $H$, their product is in $H$. In the quotient $G/H$, this becomes the product of every two non-identity element is the identity. Clearly the only group with this property has order $2$ and so $|G:H|=2$.
(See for example Theorem 3 in https://ac.els-cdn.com/0095895671900190/1-s2.0-0095895671900190-main.pdf?_tid=4f899bdc-530d-481d-8a74-d64fc2bd0f37&acdnat=1536106134_64e418ca1bb4613f97a0ace433875e8b)
Note that we have proved something stronger: either $G$ is abelian, or $H$ is an abelian subgroup of index $2$ in $G$.
By the way, the nonabelian groups with this property are called generalised dicyclic groups (see https://en.wikipedia.org/wiki/Dicyclic_group#Generalizations). They come up in a variety of settings.
