2
$\begingroup$

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?

$\endgroup$
0
2
$\begingroup$

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 $G$.

$\endgroup$
6
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.