1. ## Group theory problem

Let G be a finite group and let I ={g in G: g^2 = e} \ {e} be its subset of involutions. Show that G is abelian if card(I) => (3/4)card(G).

I don't really know how to proceed with this problem and to make use of 3/4. I've thought about group actions but my algebra is kind of rusty right now. Any help on how to proceed with the problem is highly appreciated. Thanks.

2. Originally Posted by playa007
Let G be a finite group and let I ={g in G: g^2 = e} \ {e} be its subset of involutions. Show that G is abelian if card(I) => (3/4)card(G).

I don't really know how to proceed with this problem and to make use of 3/4. I've thought about group actions but my algebra is kind of rusty right now. Any help on how to proceed with the problem is highly appreciated. Thanks.
Hi playa,

Let $\displaystyle x\in I$, $\displaystyle y \in I$. We first show that $\displaystyle xy\in I\cup\{e\}$.
Since $\displaystyle G$ is finite, map $\displaystyle a\rightarrow at$ from $\displaystyle G$ to $\displaystyle G$ is a bijection for every $\displaystyle t\in G$.

Thus, because $\displaystyle |I|\ge \frac{3}{4}|G|$, we have $\displaystyle |\{s\in G: sx\in I\}|\ge \frac{3}{4}|G|$ and $\displaystyle |\{s\in G: sy\in I\}|\ge \frac{3}{4}|G|$ and $\displaystyle |\{s\in G: sxy\in I\}|\ge \frac{3}{4}|G|$.

It follows that $\displaystyle |\{s\in G: sx\in I\}\cap \{s\in G: sy\in I\}\cap \{s\in G: sxy\in I\}|\ge \frac{1}{4}|G|$.

Further, since $\displaystyle e \not \in I$, we have $\displaystyle |\{s\in G: ss=e\}|=|I|+1\ge \frac{3}{4}|G|+1$.

So there must exist at least one $\displaystyle t\in G$ such that $\displaystyle t\in \{s\in G: sx\in I\}\cap \{s\in G: sy\in I\}\cap \{s\in G: sxy\in I\} \cap \{s\in G: ss=e\}$.

We see this $\displaystyle t$ has many desirable properties, among others, it commutes with $\displaystyle x$ and with $\displaystyle y$.

So we get $\displaystyle xyxy=ttxyxy=txtyxy=txytxy=e$, thus $\displaystyle xy\in I\cup\{e\}$.

In fact we've proved that $\displaystyle I \cup \{e\}$ is a subgroup of $\displaystyle G$.
But since $\displaystyle I$ is too fatty, $\displaystyle I \cup \{e\}=G$.

Since all elements of $\displaystyle G$ are involutions, $\displaystyle G$ must be abelian ($\displaystyle uv=uv(uvvu)=(uvuv)vu=vu$).