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.
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Sign up to join this communityLet $H\leq G$ 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.
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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.
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 $G$.
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 $H=\langle H\cup aHa^{-1}\rangle$.