Question

Let G be an abelian group.

Show that $\mathit{H}\mathbf{=}\mathbf{\text{\hspace{0.17em}}}\mathbf{\left\{}{\mathbf{x}}^{\mathbf{2}}\mathbf{|}\mathbf{x}\mathbf{\in }\mathbf{G}\mathbf{\right\}}$is a subgroup of G.

Short answer

It is proved that,$H=\text{\hspace{0.17em}}\left\{x^2|x\in G\right\}$ is a subgroup of G.

Step-by-step solution

Properties of abelian Group and Theorem 7.11

Properties of abelian Group

An abelian group is a group whose elements always hold five properties, as listed below:

1. Closer
2. Associative
3. Identity element
4. Inverse element
5. Commutative

Theorem 7.11

A nonempty subset H of a group Gis a subgroup of G provided that:

1. if$\mathit{a}\mathbf{,}\mathit{b}\mathbf{\in }\mathit{H}$, then$\mathit{a}\mathit{b}\mathbf{\in }\mathit{H}$.
2. if$\mathit{a}\mathbf{\in }\mathit{H}\mathbf{,}$then${\mathit{a}}^{\mathbf{-}\mathbf{1}}\mathbf{\in }\mathit{H}$.

Proving that H=​ {x2 | x∈ G}is a subgroup of G

Suppose.$x,y\in G$

Then,$x^2,y^2\in H$ and simplify it as:

$x^2{y}^2={\left(xy\right)}^2$

This equality is true because G is an abelian group.

Therefore,$\left(xy\right)\in H$ .

Now, for any ,$x\in G$ .$x^2\in H$

Then,${\left(x^2\right)}^{-1}={\left(x^{-1}\right)}^2$ .

This equality is also true because G is an abelian group.

Therefore,$x^{-1}\in H$ .

Since both the conditions are satisfied,$H=\text{\hspace{0.17em}}\{x^2:x\in G\}$ is a subgroup of G.

Therefore, it is proved that $H=\text{\hspace{0.17em}}\{x^2:x\in G\}$is a subgroup of G.