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Inequality on groups

Let \mathcal{G} be a finite group and suppose that \mathcal{H} , \mathcal{K} are two subgroups of \mathcal{G} such that \mathcal{H} \neq \mathcal{G} and \mathcal{K} \neq \mathcal{G}. Show that

    \[\left|\mathcal{H} \cup \mathcal{K} \right| \leq \frac{3}{4} \left| \mathcal{G} \right|\]

Solution

Recall that \displaystyle |\mathcal{H}\mathcal{K}|=\frac{|\mathcal{H}||\mathcal{K}|}{|\mathcal{H} \cap \mathcal{K}|} and thus \displaystyle \frac{|\mathcal{H}||\mathcal{K}|}{|\mathcal{H} \cap \mathcal{K}|} \leq |\mathcal{G}|. Hence \displaystyle |\mathcal{H} \cap \mathcal{K}| \ge \frac{|\mathcal{H}| |\mathcal{K}|}{|\mathcal{G}|} and so

(1)   \begin{align*} |\mathcal{H} \cup \mathcal{K}|&=|\mathcal{H}|+|\mathcal{K}|-|\mathcal{H} \cap \mathcal{K}| \\ &\leq |\mathcal{H}|+|\mathcal{K}|-\frac{|\mathcal{H}| |\mathcal{K}|}{|\mathcal{G}|} \\ & =(a+b-ab)|\mathcal{G}|  \end{align*}

where \displaystyle a=\frac{|\mathcal{H}|}{|\mathcal{G}|} and \displaystyle b=\frac{|\mathcal{\mathcal{K}}|}{|\mathcal{G}|}.

Now, since \mathcal{H} \neq \mathcal{G} and \mathcal{K} \neq \mathcal{G} we have [\mathcal{G}: \mathcal{H}] \geq 2 and [\mathcal{G}:\mathcal{K}] \geq 2 that is a \leq \frac{1}{2} and b \leq \frac{1}{2}. So if we let a'=1-2a and b'=1-2b then a', b' \geq 0 and thus

    \[a+b-ab=\frac{3}{4}-\frac{a'+b'+a'b'}{4} \leq \frac{3}{4}\]

due to (1).

The exercise along its solution have been migrated from here .

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