Question

For each $a\in G$ prove that$a^{-1}Na\subseteq N$ and apply Theorem 8.11.: [Hint: If $a\in N$and$n\in N$,$a^{-1}na$ is either in N or in Na by part (a). Show that the latter possibility leads to a contradiction

Short answer

It is proved that $a^{-1}Na\subseteq N$.

Step-by-step solution

Important Theorem

Theorem 8.11: The following conditions on a subgroup of a group 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. $aN{a}^{-1}\subseteq N$ for every $a\in G$ .

Proving that a-1Na⊆N

Let a be an arbitrary element in G,$a\in G$

We know that the index of N is two; therefore, it must have two cosets. Its right cosset beingNa and its left coset being aN

Since N is a subgroup of G, therefore:

$G=N\cup Na=Na\cup N$

This implies that either Na=N or Na=aN.

Since in ‘part (a)’ of the question, we already proved that N and Na do not have any common elements. So, $N\ne Na$.

Therefore, Na=aN and as a corollary$a^{-1}na=N\Rightarrow a^{-1}na\subseteq N$.

It is proved that $a^{-1}Na\subseteq N$.