$G\leqslant GL(n,\mathbb R)\cap\operatorname{Sym}(n,\mathbb R)$ and $|G|=m<\infty$. Prove that $G\cong (\mathbb Z/2\mathbb Z)^k$ for some $k\ge 0$.

Question

Let $G$ be a finite subgroup of the group of real $n\times n$ matrices with nonzero determinant such that all elements of $G$ are symmetric matrices. Prove that $G$ is isomorphic to $(\mathbb Z/2\mathbb Z)^k$ for some $k\ge 0$.

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My attempt:

Since $G\leqslant GL(n,\mathbb R)\cap\operatorname{Sym}(n,\mathbb R)$ and $|G|=m<\infty$, we have $\forall g\in G$, $g^m=I$ which implies that the only possible eigenvalues of $g$ are $\pm 1$.

Moreover, we need to show that $g^2=I$ and $g_1g_2=g_2g_1$ for every $g_1,g_2\in G$, which will guarantee that $G$ is a finite abelian group with the only orders of elements in $G$ is $1$ or $2$. Consequently, we can conclude that $G\cong (\mathbb Z/2\mathbb Z)^k$ for some $k\ge 0$.

Indeed, suppose $PgP^{-1}$ is in its Jordan for some $P\in\ GL(n,\mathbb R)$ then we can easily conclude that $g^2=I$. But I am stuck at showing the second fact. Any help? Thanks.

Answer 1

It is an exercise in first courses in group theory to prove $g^2=e$ for all $g\in G$ implies $gh=hg$ for all pairs of elements $g,h\in G$. The way to do it is expand out $(gh)^2=g^2h^2$ and cancel $g,h$ on the sides. (So, actually, in general it's sufficient for $x\mapsto x^2$ to be a homomorphism for $G$ to be commutative.)

Answer 2

Assume we have $g^2 = I$, that is: $g = g^{-1}$.

Then, for all $(g_1, g_2) \in G^2$, we have:

\begin{equation*} g_1 g_2 = (g_1 g_2)^{-1} = g_2^{-1} g_1^{-1} = g_2 g_1 \end{equation*}

Thus, $G$ is commutative.