Question

If G is a abelian group, prove that $\mathbf{pG}\mathbf{=}\left\{\mathrm{px}Ix\in G\right\}$ is a subgroup of G .

Short answer

It is proved that, $\mathrm{pG}=\left\{\mathrm{px}\mathrm{I}\mathrm{x}\in \mathrm{G}\right\}$ is a subgroup ofG.

Step-by-step solution

Finite Abelian Group

Every finite abelian group G is the direct sum of cyclic groups, each of prime power orders.

Group G and a subgroup  H

Let G be an additive group and H be a subset of G.

Then, H is called a subgroup if $\mathrm{a},\mathrm{b}\in \mathrm{H}$implies $\mathrm{a}+{\mathrm{b}}^{-1}\in \mathrm{H}\mathrm{or}\mathrm{a}-\mathrm{b}\in \mathrm{H}$or .

Given that G is an abelian group and $\mathrm{pG}=\left\{\mathrm{px}\mathrm{I}\mathrm{x}\in \mathrm{G}\right\}$.

pGis a subgroup

Let ${\mathrm{px}}_1,{\mathrm{px}}_2\in \mathrm{pG}$where, .

Since G is an abelian group, ${\mathrm{x}}_1-{\mathrm{x}}_2\in \mathrm{G}$which implies as:

$$\begin{gathered} {\mathrm{px}}_1-{\mathrm{px}}_2=\mathrm{p}\left({\mathrm{x}}_1-{\mathrm{x}}_2\right) \\ \in \mathrm{pG} \end{gathered}$$

Thus, ${\mathrm{px}}_1,{\mathrm{px}}_2\in \mathrm{pG}$implies${\mathrm{px}}_1-{\mathrm{px}}_2\in \mathrm{pG}$

Therefore, $\mathrm{pG}=\left\{\mathrm{px}\mathrm{I}\mathrm{x}\in \mathrm{G}\right\}$ is a subgroup of G.