Question

(a) Let and be subgroups of an abelian group G and let$\mathit{H}\mathit{K}\mathbf{=}\left\{ab|a\in H,b\in K\right\}$.Prove thatHK is a subgroup ofG.

Short answer

Itis proved that HK is a subgroup of G.

Step-by-step solution

Condition for abelian and non-abelian group

Group G is said to be abelian if ab=ba for all $a,b\in G$, and group G is said to be non-abelian if $ab\ne ba$for all $a,b\in G$.

Show that HK is a subgroup of G

Consider $HK=\left\{ab|a\in H,b\in K\right\}$.

Consider two elements $h_1{k}_1$and $h_2{k}_2$in this set. Also, consider their product and the following manipulations on it:

$\left(h_1{k}_1\right)\left(h_2{k}_2\right)=\left(h_1{h}_2\right)\left(k_1{k}_2\right)$

Since the elements$h_1$and $h_2$belong to the group $H$, so the elements $k_1$and $k_2$belong to the group $K$, and the products $h_1{h}_2$and $k_1{k}_2$belong to the set $HK$.

Consider the inverse of the element $h_1{k}_1$and apply the following manipulations on it:

$$\begin{gathered} {\left(h_1{k}_1\right)}^{-1}={k_1}^{-1}{h_1}^{-1} \\ ={h_1}^{-1}{k_1}^{-1} \end{gathered}$$

Since the element $h_1$belongs to the group Hand element $k_1$belongs to the group K, the elements ${h_1}^{-1}$and ${k_1}^{-1}$belong to the groups H and K respectively.

Therefore, the product ${h_1}^{-1}{k_1}^{-1}$belongs to the set HK . Thus, this set is a subgroup of G .

Hence, it is proved that HK is a subgroup of G.