Let $G$ be a group of order $n$ where every non identity element is of order $2$. Then, ${\rm Aut}(G) $ is isomorphic to $S_{n-1}$

Question

Let $G$ be a group of order $n$ where every non identity element is of order $2$. Then, ${\rm Aut}(G) $ is isomorphic to $S_{n-1}$. Is this true?

I read that ${\rm Aut}(K_4)$ is isomorphic to $S_3$. We can see this because of the permutations of the non identity elements, all of which have order $2$. So for each non identity element, we have $3$ choices for it's image. Proceeding recursively, we get $3!$ and ${\rm Aut}(K_4)$ is isomorphic to $S_3$.

Can we further generalize this idea to groups in which all non identity elements are of order $2$?

Thanks in advance!

Answer 1

A group in which every non-identity element has order $2$ is abelian: to see this, note that $a^2=1$ for all $a$, implies that $a = a^{-1}$ for all $a$, so that $$ab = (a^{-1})^{-1}(b^{-1})^{-1} = (b^{-1}a^{-1})^{-1} =b^{-1}a^{-1}= ba$$

for all $a$ and $b$. It follows that any group, $G$, in which every non-identity element has order $2$ is a vector space over the field $\Bbb{F}_2$ with $2$ elements, and that its group automorphisms are the invertible linear transformations. Hence, if $G$ is finite of order $n$, then $n = 2^k$ for some $k$ and the automorphism group of $G$ is isomorphic to $\mathrm{GL}_k(\Bbb{F}_2)$. The formula for $|\mathrm{GL}_k(\Bbb{F}_2)|$ given in the first answer to this question about the order of these groups shows that this automorphism group cannot be isomorphic to a symmetric group unless $n \le 4$.

Answer 2

Note that such a group is Abelian, since we have $(a + b) + (a + b) = 0$ and also $a + b + b + a = 0$ and thus $a + b = b + a$.

By the fundamental theorem of finitely generated Abelian groups, the group must be isomorphic to $\mathbb{Z}_2^k$ for some $k$.

Now consider the fact that a group homomorphism between two $\mathbb{Z}_2$ vector spaces is exactly a linear map. Then we are looking for the group of all linear isomorphisms $\mathbb{Z}_2^k \to \mathbb{Z}_2^k$.

I claim that there are $\prod\limits_{i = 0}^{k - 1} (2^k - 2^i)$ of them.

Indeed, a linear map from $\mathbb{Z}_2^k$ to $\mathbb{Z}_2^k$ amounts to picking a value of $\mathbb{Z}_2^k$ for each basis vector to map to.

If we wish to pick an invertible linear map, we must send the basis vectors to a set of linearly independent vectors.

When we pick the $i + 1$th vector, we've already picked $i$ vectors. So we cannot pick any vector in the vector space spanned by the first $i$ vectors, which has size $2^i$. But we can pick any of the other vectors. So we have $2^k - 2^i$ choices for picking where to send the $i + 1$th vector, thus showing that there are a total of $\prod\limits_{i = 0}^{k - 1} (2^k - 2^i)$ such maps.

Now note that $2^n = k$. So $|S_{n - 1}| = |S_{2^k - 1}| = (2^k - 1)!$.

Now both $(2^k - 1)$ and $\prod\limits_{i = 0}^{k - 1} 2^k - 2^i$ are products of a set of integers less than $2^k$. But the second product is only the product of $k$ such numbers, while the other is the product of all $2^k - 1$ numbers. Thus, we see that for $k > 2$, we have $|Aut(\mathbb{Z}_2^k)| > |S_{2^k - 1}|$. Therefore, the two groups are usually not isomorphic.

Answer 3

Let $a$ and $b$ be two distinct non-identity elements of $G$. If $\tau$ is an automorphism of $G$ with $\tau(a)=a$ and $\tau(b)=b$, then we must have $\tau(ab)=ab$.

If $n>4$, then there exists an element $c \in G \setminus \{e,a,b,ab\}$. In that case there is no automorphism of $G$ with $\tau(a)=a$, $\tau(b)=b$, and $\tau(ab)=c$, so not all permutations of $G \setminus \{e\}$ are possible, and hence $|{\rm Aut}(G)| < |S_{n-1}|$.