6
$\begingroup$

Let $H\leq G$ be groups, and $a\in G$ so that $\langle H,a\rangle=G$. Does it follows that $\langle H\cup aHa^{-1}\rangle$ is a normal subgroup of $G$?

My hope is that this is true, and my guess is that it is not.

It might be easy.

$\endgroup$

closed as off-topic by YCor, Pace Nielsen, Chris Godsil, Jeremy Rickard, Dima Pasechnik Aug 19 at 6:20

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "MathOverflow is for mathematicians to ask each other questions about their research. See Math.StackExchange to ask general questions in mathematics." – Chris Godsil, Jeremy Rickard, Dima Pasechnik
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 5
    $\begingroup$ An obvious counterexample is when $G=H\wr\langle a\rangle$ with $a$ of order $\ge 3$ (this gives a counterexample of order $24$ with $H$ of order $2$, $a$ of order $3$). $\endgroup$ – YCor Aug 17 at 17:16
  • 1
    $\begingroup$ Thanks, very nice. So there is even a finite counterexample. I don't know why I didn't see this. $\endgroup$ – Bertalan Bodor Aug 17 at 19:44
4
$\begingroup$

Too long for a comment. The answer is no.

Just take the free product $G=H*\langle a \rangle$, where $a\notin H$ is of infinite order. By definition, $H$ and $a$ generates $G$. Then $\langle H\cup aHa^{-1}\rangle $ is not normal.

For example, letting $h\in H$, $g:=a^2ha^{-2}\notin \langle H\cup aHa^{-1}\rangle$. Indeed by contradiction, write $g=h_1...h_n$, where $h_k\in H$ or $h_k\in aHa^{-1}$. Then, the first letter needs to be an $a$ so that $h_1\in aHa^{-1}$ but the second letter of $h_1$ is in $H$, whereas the second letter of $g$ is again $a$.

It seems to me that the problem actually comes from the powers of $a$. For example, assume that $a^2\in H$, then, the answer is yes. Indeed you then have that $G/H$ is of order 2, so that $H$ is normal in $G$ and so $\langle H\cup aHa^{-1}\rangle =H$ is indeed normal. So if you have a particular group in mind, maybe you should first look carefully as the relations between powers of $a$ and $H$.

Also quick comment: try to find more suited titles.

$\endgroup$
  • $\begingroup$ Thanks, very nice. $\endgroup$ – Bertalan Bodor Aug 17 at 19:34
3
$\begingroup$

As other have said, the answer is no in general. However if $a$ has finite order $n$, then it is true that $\langle H \cup aHa^{-1} \cup \ldots \cup a^{n-1}Ha^{-(n-1)} \rangle$ is a normal subgroup of $G$. This is because the group in question is normalized by $H$ (since it contains $H$ ) and is normalized by $a$, since the given generating set is stable under conjugation by $a$ ( and hence the group it generates is stable under conjugation by $a$). Hence the given group is normalized by $\langle H, a \rangle = G$, so is a normal subgroup of $G$.

$\endgroup$
2
$\begingroup$

Work in $GL_2(\mathbb{Q})$. Let $H$ be the subgroup generated by $$ \left(\begin{array} 01 & 1\\ 0 & 1\end{array}\right). $$ Let $G=\langle H,a\rangle$ where $$ a=\left(\begin{array} 02 & 0\\ 0 & 1\end{array}\right). $$ Then $aHa^{-1}$ is a proper subgroup of $H$. In particular, $H$ is not normal in $G$, and $H=\langle H\cup aHa^{-1}\rangle$.

$\endgroup$

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