# Show that $\bigcap_{a \in G} a H a^{-1}$ is a normal subgroup of $G$.

I am reading "An introduction to algebraic systems" by Kazuo Matsuzaka.

In this book, there is the following problem.

I think this problem is easy but a little abstract for me.

Let $$G$$ be a group.
Let $$H$$ be a subgroup of $$G$$.
Let $$N = \bigcap_{a \in G} a H a^{-1}$$.

(1)
Show that $$N$$ is a normal subgroup of $$G$$.

(2)
Show that if $$M$$ is a normal subgroup of $$G$$ and $$M \subset H$$, then, $$M \subset N$$.

(1)
Let $$g_1, g_2$$ be arbitrary elements of $$G$$.
Let $$n$$ be an arbitrary element of $$N$$.
Because $$g_1^{-1} g_2 \in G$$ and $$n \in N$$, so $$n \in g_1^{-1} g_2 H (g_1^{-1} g_2)^{-1}$$.
So, $$n = g_1^{-1} g_2 h (g_1^{-1} g_2)^{-1}$$ for some $$h \in H$$.
Then, $$g_1 n g_1^{-1} = g_1 g_1^{-1} g_2 h (g_1^{-1} g_2)^{-1} g_1^{-1} = g_1 g_1^{-1} g_2 h g_2^{-1} g_1 g_1^{-1} = g_2 h g_2^{-1} \in g_2 H g_2^{-1}$$.
$$g_2$$ was an arbitrary element of $$G$$.
So, $$g_1 n g_1^{-1} \in N$$.
So, $$N$$ is a normal subgroup of $$G$$.

(2)
Let $$m$$ be an arbitrary element of $$M$$.
Let $$g_3$$ be an arbitrary element of $$G$$.
Then, $$g_3^{-1} m (g_3^{-1})^{-1} = g_3^{-1} m g_3 \in M \subset H$$.
$$m = g_3 (g_3^{-1} m g_3) g_3^{-1} \in g_3 H g_3^{-1}$$.
$$g_3$$ was an arbitrary element of $$G$$.
So, $$m \in N$$.

For (1): N is normal iff $$xNx^{-1} = N \forall x \in G$$ . But $$xNx^{-1} = \cap_{a \in G} (xa)H(xa)^{-1}$$ and just note that the function $$a \to xa$$ permutes the elements of G.
For (2): $$M \subset H$$ and M normal implies that $$M = \cap_{a \in G} aMa^{-1} \subset \cap_{a \in G} aHa^{-1} =N$$, as needed.