Question

If $\mathbf{N}\mathbf{\ne }<e>$is a normal subgroup of G and $\left|G\right|\mathbf{=}{\mathbf{p}}^{\mathbf{n}}$, prove that $\mathbf{N}\mathbf{\cap }\mathbf{\mathbb{Z}}\left(G\right)\mathbf{\ne }<e>$. [Hint: Exercise 14(c) may be helpful.]

Short answer

It has been proved that,.

Step-by-step solution

Given that

It is given that is a normal subgroup of G.

Also, $\left|\mathrm{G}\right|={\mathrm{p}}^{\mathrm{n}}$.

Use Exercise 14(c)

Firstly, N is a subgroup of a p-group, so it must have order p^m for some m.

Since N is normal, therefore, N is the union of the conjugacy classes C_i contained in N.

Since G is a p-group, each C_i has order pⁱ for some i.

Prove that N∩ℤ(G)≠<e>

Now, $e\in N$and $C\left(e\right)=\left\{e\right\}$has only one element, $\left|\mathrm{C}\left(\mathrm{e}\right)={\mathrm{p}}^0=-1\right|$. But since $\left|N\right|=p^m$, the order of the conjugacy classes are either 1 or divisible by pC(e) and cannot be the only conjugacy class of size 1.

If $a\in N$is such that $a\ne e$and $C\left(a\right)=\left\{a\right\}$this means that for all $g\in G$there is $g^{-1}ag=a$and so $a\in Z\left(G\right)$.

Thus, it can be concluded that .