# Subgroup generated by a subgroup and a conjugate of it [closed]

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.

• 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 $3$). – YCor Aug 17 '19 at 17:16
• Thanks, very nice. So there is even a finite counterexample. I don't know why I didn't see this. – Iras Aug 17 '19 at 19:44

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.

• Thanks, very nice. – Iras Aug 17 '19 at 19:34

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 $$H=\langle H\cup aHa^{-1}\rangle$$.

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 $$G$$.