5
$\begingroup$

Let $H_1,H_2$ be two distinct subgroups of finite group $G$ each of order $2$. Let $H$ be the smallest subgroup containing $H_1$ and $H_2$ . Then is it necessary that order of $H$ is amongst $2,4,8?$

I think that this is not true if we consider $S_3$ and $H_1=\{(1),(12)\}$ , $H_2=\{(1),(23)\}$ then smallest subgroup will be $S_3$ itself!

$\endgroup$
  • 5
    $\begingroup$ Yes, this is a valid counterexample to the statement. $\endgroup$ – Brian Dec 12 '13 at 4:22
  • 2
    $\begingroup$ As a matter of fact, given any positive integer $n$, there exists a group in which there are elements $x$ and $y$, each of order $2$, whose product $xy$ has order $n$. $\endgroup$ – Alexander Gruber Dec 12 '13 at 9:30
  • $\begingroup$ Alexander, great remark! One up from me! $\endgroup$ – Nicky Hekster Dec 12 '13 at 10:39
3
$\begingroup$

If $H_1$ or $H_2$ is normal, then your statement is true, in fact in that case $H=H_1H_2$ and $|H_1H_2|=4$. In general, the set $H_1H_2$ has 4 elements, since $|H_1H_2|=|H_1||H_2|/|H_1 \cap H_2|=2.2/1=4$, and of course $H_1H_2 \subseteq H$. So $H$ has at least 4 elements.

$\endgroup$
0
$\begingroup$

$H_1$={$1,a$}, $H_2=${$1,b$}. As the subgroups are distinct $a$ not equal to $b$. |$H_1\cup H_2|=3$. Note that $a,b$ are self inverse elements in $G$. Again if $H$ is the smallest group containing $H_1\cup H_2$ then $ab \in H$. Case 1. If $G$ abelian then $ab$ is self inverse element. so |H|=4. Case 2. If $G$ is not abelian $b^-1a^-1\in H$ . So |H|=5.

$\endgroup$

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.