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?


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 $G$.

  • $\begingroup$ I think that your solution is right. However, to be completely formal the assertion "Since $H_2\not\subseteq H_1$" would need few words of justifications. Assume $H_2\subseteq H_1$. Then $G\setminus H_1= aH_1= H_3$, which is absurd since $aH_1$ is not a subgroup. $\endgroup$ – rafforaffo Feb 3 '15 at 15:36
  • $\begingroup$ I dont understand this part $H_1∩H_2$ is of index $2$ in $H_2$ $\endgroup$ – corcia candy Feb 3 '15 at 15:37
  • $\begingroup$ @rafforaffo Well, if $H_2\subseteq H_1$, then $G= H_1\cup H_3$ and this is possible iff $H_1\subseteq H_3$ or $H_3\subseteq H_1$ (the union of two subgroups is a subgroup iff one of them is contained in the other). Hence, one of them wouldn't be proper. $\endgroup$ – SMM Feb 3 '15 at 15:41
  • 1
    $\begingroup$ @corciacandy Use The Second Isomorphism theorem. $H_1$ is a normal subgroup of $G$ and $G=H_1H_2$, so $G/H_1= H_1H_2/H_1\cong H_2/(H_1\cap H_2)$. Hence $|H_2:H_1\cap H_2|= |G:H_1|= 2$. $\endgroup$ – SMM Feb 3 '15 at 15:44
  • $\begingroup$ how $G=H_1H_2$? I dont understand @SMM $\endgroup$ – corcia candy Feb 3 '15 at 15:48

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