Involutions and Abelian Groups

Solution 1:

(1) This is the Principle of Inclusion-Exclusion from basic combinatorics.

(2) Set $ A := I(G) $. Let $ x \in I(G) $. As left-multiplication by a group element is an injective operation on the group, we see that $ B := xA $ has the same cardinality as $ A $. Hence, $$ |A| = |B| \geq \frac{3}{4} |G|. $$ Using (1), we thus obtain \begin{align} |I(G) \cap x[I(G)]| &= |A \cap B| \\ &\geq |A| + |B| - |G| \\ &\geq \frac{3}{4} |G| + \frac{3}{4} |G| - |G| \\ &= \frac{1}{2} |G|. \end{align}

(3) Let $ x \in I(G) $, and note that $ 1_{G} \in {C_{G}}(x) $. Suppose that $ g \in I(G) $ satisfies $ xg \in I(G) $. Then \begin{align} xg &= (xg)^{-1} \quad (\text{As $ (xg)^{2} = 1_{G} $.}) \\ &= g^{-1} x^{-1} \\ &= gx, \quad (\text{As $ g^{2} = 1_{G} = x^{2} $.}) \end{align} which yields $ g \in {C_{G}}(x) $. Hence, $$ \{ 1_{G} \} \cup \{ g \in I(G) ~|~ gx \in I(G) \} \subseteq {C_{G}}(x). $$

(4) Let $ x \in I(G) $. Observe that \begin{align} |{C_{G}}(x)| &\geq |\{ 1_{G} \} \cup \{ g \in I(G) ~|~ gx \in I(G) \}| \quad (\text{By (3).}) \\ &= |\{ 1_{G} \} \cup (I(G) \cap x^{-1} [I(G)])| \\ &= |\{ 1_{G} \}| + |I(G) \cap x^{-1} [I(G)]| \quad (\text{As $ 1_{G} \notin I(G) $.}) \\ &\geq 1 + \frac{1}{2} |G| \quad (\text{Applying (2) to $ x^{-1} $, which is an element of $ I(G) $.}) \\ &> \frac{1}{2} |G|. \end{align} Hence, $ [G:{C_{G}}(x)] < 2 $, which yields $ {C_{G}}(x) = G $, or equivalently, $ x \in Z(G) $.

As $ x \in I(G) $ is arbitrary, we have $ I(G) \subseteq Z(G) $. Then as $ |I(G)| \geq \dfrac{3}{4} |G| $, it follows that $$ [G:Z(G)] \leq \frac{4}{3} < 2. $$ Therefore, $ Z(G) = G $.

Conclusion: $ G $ is an abelian group.