Question

Let G be an abelian group.

Show that$\mathit{K}\mathbf{=}\mathbf{\left\{}\mathbf{a}\mathbf{\in }\mathbf{G}\mathbf{|}\mathbf{\left|}\mathbf{a}\mathbf{\right|}\mathbf{\le }\mathbf{2}\mathbf{\right\}}$ is a subgroup of G.

Short answer

It is provedthat$K=\left\{a\in G|\left|a\right|\le 2\right\}$, 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 thatK = {a∈G |  |a|≤2} is a subgroup of G

Suppose $a,b\in G$are two arbitrary elements.

Now, we have two prove that if$a,b\in \text{\hspace{0.17em}}K$ then$ab\in k$ .

Since $\left|a\right|\le 2$, for $a,b\in \text{\hspace{0.17em}}K$any$a,b\in \text{\hspace{0.17em}}K$ , ${\left(a{b}^{-1}\right)}^2=a^2{b}^{-2}$.

This is true because G is an abelian group.

Therefore,$a^2{b}^{-2}\in k$ .

Again, we have:

$\begin{array}{c}{\left(a{b}^{-1}\right)}^2=a^2{b}^{-2}\\ ={\left(b^{-1}a\right)}^{-2}\end{array}$

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

This shows that inverse also belongs to K.

Therefore, second condition of subgroup is true as well.

Hence, it is provedthat $K=\left\{a\in G|\left|a\right|\le 2\right\}$is a subgroup of G.