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Suppose that $ G $ is a finite group where at least three-fourths of the elements are involutions, i.e., $$ |I(G)| \geq \frac{3}{4} |G|. $$ (Here, $ I(G) $ denotes the set of all involutions of $ G $, where an involution of $ G $ is defined as a group element of order $ 2 $.)

(1) Suppose that $ A $ and $ B $ are subsets of a finite set $ S $. Show that $ |A \cap B| \geq |A| + |B| - |S| $.

(2) Let $ x \in I(G) $. Show that $ |I(G) \cap x[I(G)]| \geq \dfrac{1}{2} |G| $.

(3) Let $ x \in I(G) $. Show that $ \{ 1_{G} \} \cup \{ g \in I(G) ~|~ xg \in I(G) \} \subseteq {C_{G}}(x) $.

(4) Use your answers to (1)-(3) to conclude that $ G $ is an abelian group.

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This is homework, I guess? Can you tell us from what text it comes, what you have tried, and so on/ –  Mariano Suárez-Alvarez Jan 13 '13 at 19:54
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@user57149: Welcome to MSE! Can you please format your questios using MathJax? Regards –  Amzoti Jan 13 '13 at 19:58
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@user57149 I like this question very much; however, please improve your etiquette. We like to see what progress you have made on a problem so we know what you need help with. Also, many users get upset when questions are phrased as commands. With that said, welcome to MSE. Be sure to accept Haskell's answer. :) –  Alexander Gruber Jan 13 '13 at 21:10
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1 Answer

(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.

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