Question

If both N and Kare normal subgroups of G, prove that NK is normal.

Short answer

It is proved that if both N and K are normal, then NK is also normal.

Step-by-step solution

Required Theorem

Theorem 8.11: The following conditions on a subgroup N of a group G are equivalent:

1. Nis a normal subgroup of G.
2. $a^{-1}Na\subseteq N$for every $a\in G$, Where $a^{-1}Na\subseteq N$$\left\{a^{-1}Na| n\in N\right\}$
3. $aN{a}^{-1}\subseteq N$for every $a\in G$ , Where $aN{a}^{-1}\subseteq N$$\left\{aN{a}^{-1}| n\in N\right\}$
4. $a^{-1}Na\subseteq N$for every $a\in G$ .
5. $a^{-1}Na\subseteq N$for every $a\in G$ .

Proving that if both N and K normal, then NK  is also normal.

It is given that N and K are normal subgroups of G .

Therefore, from the theorem 8.11, we know that

$an{a}^{-1}\subseteq N$and $ak{a}^{-1}\subseteq K$ (such that $n\in N$, $k\in K$ and $a\in G$)

Now, if NK is normal, then for any $nk\in NK$and $a\in G$

$\begin{array}{l}a\left(nk\right)a^{-1}\subseteq NK\\ an{a}^{-1}ak{a}^{-1}\subseteq NK\end{array}$

Since we know$an{a}^{-1}\subseteq N$and $ak{a}^{-1}\subseteq K$

Therefore, it can be concluded that $a\left(nk\right)a^{-1}\subseteq NK$

Hence, from the above result, it is proved that NK is also a normal subgroup of G .