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.
Is my answer ok?
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, $m \in N$.
Your proof is ok but your write-up is slightly overcomplicating matters. Here is a slightly shorter and perhaps more enlightening way to understand what is going on.
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.