# Prove $\langle X \rangle \triangleleft G$ iff $gXg^{-1}\subseteq \langle X \rangle$ for all $g \in G$.

$X$ is a nonempty subset of a group $G$.

Prove $\langle X \rangle \triangleleft G$ iff $gXg^{-1}\subseteq \langle X \rangle\,\forall\, g \in G$.

Where $\langle X \rangle \triangleleft G$ means $\langle X \rangle$ is normal in $G$

I believe that $\langle X \rangle \triangleleft G \implies gXg^{-1}\subseteq \langle X \rangle\,\forall\, g \in G$ is simple. i.e.

$\langle X \rangle \triangleleft G \implies g \langle X \rangle g^{-1}\subseteq \langle X \rangle \,\forall\, g\in G \implies g X g^{-1}\subseteq g \langle X \rangle g^{-1}\subseteq \langle X \rangle \,\forall\, g\in G$

But now I am trying to show that $gXg^{-1}\subseteq \langle X \rangle\,\forall\, g \in G \implies \langle X \rangle \triangleleft G$ and I am not sure how. If $X=\langle X \rangle$ then I'd be done, but I don't think I have anyway to show that. But I know I need to show $gXg^{-1}\subseteq \langle X \rangle \implies g\langle X \rangle g^{-1}\subseteq \langle X \rangle$ Can anyone point me in the right direction? Thank you

Since you have $gXg^{-1}\subseteq \langle X \rangle\,\forall g \in G$, for each $x \in \langle X \rangle$, we can express it as a finite product $\prod_{i = 1}^n x_i$ for some $x_i \in X\,\forall i = 1,\dots,n$. Then it suffices to show that $gxg^{-1} \in \langle X \rangle$. Note that $gxg^{-1}=g\left(\prod_{i = 1}^n x_i\right)g^{-1}=\prod_{i = 1}^n (gx_ig^{-1})$. For each $i = 1,\dots,n$, $gx_ig^{-1} \in \langle X \rangle$, so it's a finite product of elements of $X$, so as $\prod_{i = 1}^n (gx_ig^{-1})=gxg^{-1}$. Hence $gxg^{-1} \in \langle X \rangle$ for all $x \in \langle X \rangle$.
• I'm new to Abstract Algebra and I learnt that to show that if H is normal in G, we have to show that $gHg^{-1}=H$ for all $g \in G$. Why do you say that it suffices to show that $gxg^{-1} \in <X>$? – Icycarus Nov 17 '18 at 23:17