Question

Question:If N is a normal subgroup of a (not necessarily finite) group G and both N and $\mathit{G}\mathbf{/}\mathit{N}$are p-groups, then prove that G is a p-group.

Short answer

It is proved that G is a p-group.

Step-by-step solution

Given that

Given that N is a normal subgroup of a group G and both N and G/N are p-groups.

Proving that G is a p-group

Suppose g is an arbitrary element of G,$g\in G$.

Since$G/N$ is a p-group, its order can be given as$p^a$ , where a is an arbitrary element such that$g^{p^a}\in N$ .

Since N is also a p-group, its order can be given as$p^b$ , where b is an arbitrary element.

As we know,$g^{p^a}\in N$,$g^{p^a}$ has order $p^b$in group N.

Thus, element g has order$p^{\left(a+b\right)}$ , i.e., sum power of p.

Therefore, the element of G has the power of p which implies G is a p-group.

Thus,G is a p-group.