Question

Question: If $G$ is a nilpotent group (see Exercise 13 of Section 9.3), prove that $G$ has this property: If $m$ divides $\left|G\right|$, then $G$ has a subgroup of order $m$ .[You may assume Exercise 22.]

Short answer

It is proved that,- $G$ has a subgroup of order$m$.

Step-by-step solution

Prove G has a subgroup of order m

Given that G is a nilpotent group.

By definition, it is isomorphic to the direct product of its Sylow p-subgroups for the prime p that divides $\left|G\right|$.

Let $m=\prod _{i=1}^k{p}_i^{a_i}$be a divisor of $\left|G\right|$and suppose the order of the Sylow $p_i$-subgroup $G\left(p_i\right)$is $p_i^{{\beta }_i}$ with ${\alpha }_i\le {\beta }_i$.

By multiple application of exercise 22, we have a subgroup $H_i$of order $p_i^{{\alpha }_i}$in $G\left(p_i\right)$

We then have that $\prod _{i=1}^k{H}_i$ is a subgroup of $\prod _{i=1}^{k}G\left(p_i\right)$of order $m$, which is associated by the isomorphism $G\cong \prod _{i=1}^{k}G\left(p_i\right)$to a subgroup of $G$of order $m$.

Hence, $G$has a subgroup of order $m$.