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If $G =\langle X\rangle$, then for $N\le G$ to be a subgroup normal it is enough that $N^x = N$ for all $x\in X$, and if $G$ is finite it is enough that $N^x \le N$ for all $x \in X$.

I have tried to prove that in the infinite case, only the second condition is necessary. Have I made a mistake?

If $G$ is infinite, then each element $g$ of $G$ can be expressed as a finite quantity of products of elements of $X$. Now, if $m\in N^g$, then $m = n^g$ for some $n\in N$, then $n\in N$ and if we consider that $N^x\le N$ for all $x\in X$, since each $g$ is a finite product of a number of elements of $X$ and $n\in N$, then we can form $n^g$ as follows. We list the factors of $g$ in the following way $g = x_1 ... x_n$ in such a way that each of these factors belongs to $X$, then $n^{x_1}\in N$, then $n^{x_1x_2}\in N$ and thus we form $n^{x_1 ... x_n} = n^g\in N$. Therefore, $N^g\subseteq N$ (repeating the same procedure for each $g\in G$). Then $N$ is a normal subgroup.

Shaun
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MenuPrincipal
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    "can be expressed as a finite quantity of products of elements of $X$". It is not true for infinite groups: you also need $X^{-1}$. In a finite group or, more generally, torsion groups for every $x$ there exists a natural $n$ such that $x^n=x^{-1}$ so your argument works for finite groups. But for infinite groups you also need $N^{x^{-1}}\le N$ for every $x\in X$. Together with $N^x\le N$ this gives $N^x=N$. – markvs Jan 11 '22 at 04:14
  • You're right, thank you very much for helping me. – MenuPrincipal Jan 11 '22 at 04:25
  • To see that the result is not true as stated, look at the example here. It gives an example of a group $G$, a subgroup $H$, and an element $x$ such that $xHx^{-1}\subseteq H$ but $x^{-1}Hx\not\subseteq H$. Taking that example and replacing $G$ with $\langle H,x\rangle$ and letting $X={H,x^{-1}}$ gives you an explicit counterexample. – Arturo Magidin Jan 11 '22 at 17:14

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As noted, the result is false in the infinite setting, unless $X$ is closed under inverses, or you know that every element of $X$ is torsion; or $N$ is itself finite. The problem is that in general you cannot express $x^{-1}$, with $x\in X$, as a product of elements of $X$ unless $x$ has finite order.

For an explicit counterexample, start with the group of invertible matrices with coefficients in $\mathbb{Q}$ of the form $$\left(\begin{array}{cc} a&b\\0&c\end{array}\right),\qquad ac\neq 0.$$ Let $N$ be the subgroup of matrices with $a=c=1$, $b\in\mathbb{Z}$; and let $$y = \left(\begin{array}{cc} \frac{1}{2}&0\\0&1\end{array}\right).$$ Finally, let $G=\langle N,y\rangle$, and set $X=\{y\}\cup\{z\mid z\in N\}$.

Since $N$ is a subgroup and $$y^{-1}\left(\begin{array}{cc} 1&b\\ 0&1\end{array}\right)y = \left(\begin{array}{cc} 1&2b\\ 0&1\end{array}\right),$$ it follows that $x^{-1}Nx\subseteq N$ for every $x\in X$.

However, $$y\left(\begin{array}{cc} 1&1\\0&1\end{array}\right)y^{-1} = \left(\begin{array}{cc} \frac{1}{2}&0\\ 0&1\end{array}\right)\left(\begin{array}{cc}1&1\\0&1\end{array}\right) \left(\begin{array}{cc}2&0\\0&1\end{array}\right) = \left(\begin{array}{cc} 1&\frac{1}{2}\\ 0&1\end{array}\right)\notin N,$$ so $N$ is not normal.

Your argument fails because in general, if $x\in X$, you do not know whether you can express $x^{-1}$ as a product of elements of $X$. You can do it if you know that for every $x\in X$, either $x$ is of finite order or else $x^{-1}\in X$. If you assume that disjunction, then your argument goes through. Alternatively, if $N$ is finite then you can conclude that $N^x=N$ (because $N^x\leq N$ and both finite of the same order gives equality), hence $N^{x^{-1}}=N$ holds as well.

There are two ways to see the condition holds in the finite case. One is that in that case, the inverses of elements of $X$ can be written as products of elements of $X$ (we are in the case where every element is of finite order). Alternatively, because $|N^x|=|N|$ and both are finite, so $N^x\leq N\implies N^x=N$ and from there you are fine.

Arturo Magidin
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