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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, $m \in N$.

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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.

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    $\begingroup$ Thank you very much, for clear proof, Sorin Tirc. $\endgroup$ – tchappy ha Jan 23 '19 at 12:07

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