Question

(b) Show that part (a) may be false if G is not abelian.

Short answer

It is proved that the set is not a subgroup.

Step-by-step solution

Condition for abelian and non-abelian group

Group is $\mathit{G}$said to be abelian if $\mathit{a}\mathit{b}\mathbf{=}\mathit{b}\mathit{a}$for all$\mathit{a}\mathbf{,}\mathit{b}\mathbf{\in }\mathit{G}$ , and group $\mathit{G}$is said to be non-abelian if $\mathit{a}\mathit{b}\mathbf{\ne }\mathit{b}\mathit{a}$for all $\mathit{a}\mathbf{,}\mathit{b}\mathbf{\in }\mathit{G}$.

Show that HK is a subgroup of G

The subset $HK=\left\{ab|a\in H,b\in K\right\}$may not be a subgroup of $G$if the latter is not abelian.

Let $G$be the symmetric group $S_4$and the required subgroups are defined as follows:

$$\begin{gathered} H=\left\{\left(1\right),\left(12\right)\right\} \\ K=\left\{\left(1\right),\left(13\right)\right\} \end{gathered}$$

Obtain the set $HK$as follows:

$$\begin{gathered} HK=\left\{\left(1\right)\left(1\right),\left(1\right)\left(13\right),\left(12\right)\left(1\right),\left(12\right)\left(13\right)\right\} \\ =\left\{\left(1\right),\left(13\right),\left(12\right),\left(132\right)\right\} \end{gathered}$$

It can be observed that this set is not closed under multiplication because the product of the permutations (13) and (12) is the permutation (123), which is not in the set.

Therefore, this set is not a subgroup.