Subgroup generated by a subgroup and a conjugate of itsto
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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2$\\begingroup$ 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 14 hours ago
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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 11 hours ago
3 Answers
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.
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$\\begingroup$ Thanks, very nice. $\\endgroup$ – Bertalan Bodor 12 hours ago
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 $H=\\langle H\\cup aHa^{-1}\\rangle$.
