# If $x$ and $y$ are conjugates, and $H_1$ the smallest normal subgroup containig $x$ and $H_2$ that containing $y$. Show that $|H_1| = |H_2|$

Problem:

If $x$ and $y$ are conjugates, show that the order of the smallest normal subgroup containing $x$ equals the order of the smallest normal subgroup containing $y$.

I proved (maybe wrongly) that $H_1 = H_2$, and not that $|H_1| = |H_2|$, so I would like some guidance here.

My attempt:

Let $H_1 \subset G$ be the smallest normal group containing $x$ and $H_2 \subset G$ the smallest subgroup containing $y$.

By definition (I am going to say ''definition'', but in my lectures notes it is actually a theorem) $H_1$ is normal $\iff \big[(\forall x \in H_1)(\forall g\in G) \implies gxg^{-1}\in H_1\big] \implies y \in H_1 \implies H_1 \subset H_2$.

On the other hand $H_2$ is normal $\iff \big[(\forall y \in H_2)(\forall g\in G) \implies gyg^{-1}\in H_2\big] \implies$ $\implies x \in H_2 \implies H_2 \subset H_1$.

Thus we have that $H_1 = H_2 \implies |H_1| = |H_2|$.

You're right. Easy argument: intersection of all normal subgroups containing subset $X \subset G$ is indeed the smallest one with such property and unique (it's usually called the normal closure of a subset). So if $x = g^{-1}yg$, then $y$ lies in normal closure of $x$ and vise versa.