## Maximum number of elements of the union of two proper subgroups of a finite group

Posted: December 2, 2018 in Elementary Algebra; Problems & Solutions, Groups and Fields
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Problem 1. Let $G$ be a group and suppose that $H, K$ are two subgroups of $G.$ Show that if $G=H \cup K,$ then either $H=G$ or $K=G.$

Solution. If $H \subseteq K$ or $K \subseteq H,$ then $H \cup K=G$ gives $K=G$ or $H=G$ and we are done. Otherwise, there exist $h \in H \setminus K$ and $k \in K \setminus H.$ But then $hk \in G \setminus H \cup K,$ contradiction! $\Box$

So, as a result, if $G$ is a finite group and $H,K$ are two subgroups of $G$ with $H \ne G$ and $K \ne G,$ then $|H \cup K| \ne |G|.$ That raises this question: how large could $|H \cup K|$ get? The following problem answers this question.

Problem 2. Let $G$ be a finite group and suppose that $H, K$ are two subgroups of $G$ such that $H \ne G$ and $K \ne G.$ Show that $\displaystyle |H \cup K| \le \frac{3}{4}|G|.$

Solution. Recall that $\displaystyle |HK|=\frac{|H||K|}{|H \cap K|}$ and thus $\displaystyle \frac{|H||K|}{|H \cap K|} \le |G|.$ Hence $\displaystyle |H \cap K| \ge \frac{|H| |K|}{|G|}$ and so

\displaystyle \begin{aligned}|H \cup K|=|H|+|K|-|H \cap K| \le |H|+|K|-\frac{|H| |K|}{|G|} =(a+b-ab)|G|, \ \ \ \ \ \ \ \ \ (*)\end{aligned}

where $\displaystyle a:=\frac{|H|}{|G|}$ and $\displaystyle b:=\frac{|K|}{|G|}.$
Now, since $H \ne G$ and $K \ne G,$ we have $[G:H] \ge 2$ and $[G:K] \ge 2,$ i.e. $\displaystyle a \le \frac{1}{2}$ and $\displaystyle b \le \frac{1}{2}.$ So if we let $a':=1-2a$ and $b':=1-2b,$ then $a' \ge 0, \ b' \ge 0$ and thus

$\displaystyle a+b-ab=\frac{3}{4}-\frac{a'+b'+a'b'}{4} \le \frac{3}{4}.$

The result now follows from $(*). \ \Box$

Example 1. The upper bound $\displaystyle \frac{3}{4}|G|$ in Problem 2 cannot be improved, i.e. there exists a group $G$ and subgroups $H, K$ of $G$ such that $\displaystyle |H \cup K|=\frac{3}{4}|G|.$ An example is the Klein-four group $G=\mathbb{Z}_2 \times \mathbb{Z}_2$ and the subgroups $H:=\{(0,0), (1,0)\}$ and $K:=\{(0,0),(0,1)\}.$ Then $|G|=4$ and $\displaystyle |H \cup K|=3=\frac{3}{4}|G|.$

Example 2. We showed in Problem 1 that a group can never be equal to the union of two of its proper subgroups. But there are groups that are equal to the union of three of their proper subgroups. The smallest example, again, is the Klein-four group

$\mathbb{Z}_2 \times \mathbb{Z}_2= \{(0,0), (1,0)\} \cup \{(0,0), (0,1)\} \cup \{(0,0),(1,1)\}.$