$G=H_1 \cup H_2 \cup H_3$ index of $H_i$

Question

Let $G=H_1 \cup H_2 \cup H_3$ be a finite group, where each $H_i$ is proper subgroup of G. (I can show) $H_i \neq H_j$ where $i\neq j$

Show that each $H_i$ has index two in G

Any suggestion?

Answer 1

Set $|G|=n$. First note that at least one of $H_1,H_2,H_3$ is of index $2$: if $|G:H_i|\geq 3$ for all $i$, then $|H_i|\leq \frac{n}{3}$, and since $e\in H_1\cap H_2\cap H_3$ , then $|H_1\cup H_2\cup H_3|<n=|G|$, which gives a contradiction.

So, let $|G:H_1|=2$ and $G= H_1\sqcup aH_1$.

Set $|G:H_2|=k\geq 2$, $|H_2|= \frac{n}{k}$. Since $H_2\nsubseteq H_1$, we see that $H_1\cap H_2$ is of index $2$ in $H_2$, hence $|H_1\cap H_2|= |aH_1\cap H_2|= \frac{n}{2k}$.

Similarly, set $|G:H_3|=l\geq 2$, $|H_3|= \frac{n}{l}$, $|H_1\cap H_3|= |aH_1\cap H_3|= \frac{n}{2l}$.

Note that $aH_1= aH_1\cap G= aH_1\cap(H_1\cup H_2\cup H_3)= (aH_1\cap H_2)\cup (aH_1\cap H_3)$. Hence $$\frac{n}{2}= |aH_1|\leq |aH_1\cap H_2|+|aH_2\cap H_3|= \frac{n}{2k}+\frac{n}{2l}.$$ So $1\leq \frac{1}{k}+\frac{1}{l}$. Since $k,l\geq 2$ that is true iff $k=l=2$. Therefore, $H_2,H_3$ are also of index $2$ in $G$.