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Let $G,H$ be groups with $H\leq G$. We will prove (1) that $H\trianglelefteq G$ implies $aHa^{-1}\subseteq H$ for all $a\in G$ and (2) that if $aHa^{-1}\subseteq H$ for all $a\in G$ then $H\trianglelefteq G$.

Suppose $H\trianglelefteq G$. We will show that $aHa^{-1}\subseteq H$ for every $a\in G$. Let $a\in G$. Pick $h_1\in H$. Because $H\trianglelefteq G$ we know that $aH=Ha$ for every $a\in G$. Hence we know that there is some $h_{2}\in H$ such that $ah_{1}=h_{2}a$. By operating on both sides on the right by $a^{-1}$, the inverse of $a$ in G, we see that $ah_{1}a^{-1}=h_{2}$. Let $p=ah_1a^{-1}$. Then $p=ah_1a^{-1}=h_2\in H$ This means $aHa^{-1}\subseteq H$ for all $a\in G$.

Now suppose $aHa^{-1}\subseteq H$ for every $a\in G$. We must show $Ha=aH$ for all $a\in G$. Let $a\in G$.
$\quad$We will first show that $aH\subseteq Ha$. Let $x\in aH$. This means we pick $h_{1}\in H$ such that $x=ah_{1}$. Because $aHa^{-1}\subseteq H$ we know $xa^{-1}=ah_{1}a^{-1}=h_{2}$ for some $h_{2}\in H$. So by operating on the right by $a$ on each side of the equality we see that $x=h_{2}a\in Ha$. Thus $aH\subseteq Ha$ for all $a\in G$.
$\quad$We will now show $Ha\subseteq aH$. Pick $y\in Ha$. Then we pick $h_{3}\in H$ such that $y=h_{3}a$. Because $aHa^{-1}\subseteq H$ for every $a\in G$ we know $a^{-1}Ha\subseteq H$. This means $a^{-1}y=a^{-1}h_{3}a=h_{4}$ for some $h_{4}\in H$. It follows that $y=h_{3}a=ah_{4}\in aH$. Hence $Ha\subseteq aH$.
Because $aH\subseteq Ha$ and $Ha\subseteq aH$ for all $a\in G$ we know that $Ha=aH$ for every $a\in G$. Therefore, by definition $H\trianglelefteq G$.

We conclude that $H\trianglelefteq G$ if and only if $aHa^{-1}\subseteq H$ for all $a\in G. \blacksquare$

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