Question

Let G be an abelian group, K a fixed positive integer, and $\text{H=\hspace{0.17em}\hspace{0.17em}}\left\{\left.\text{a∈\hspace{0.17em}G\hspace{0.17em}}|\left|\text{a}\right|\text{\hspace{0.17em}\hspace{0.17em}divides\hspace{0.17em}\hspace{0.17em}k}\right\}\right.$.Prove that H is a subgroup of G.

Short answer

Answer

It is proved that H is a subgroup of G.

Step-by-step solution

Step by Step Solution: Step 1: Prove that  is closed under group operation

Let, $a,b\in H$

$$\begin{gathered} Therefore,\left|a\right||kand\left|b\right||k \\ Suppose\left[\left|a\right|,\left|b\right|\right]denotetheleastcommonmultipleof\left|a\right|and\left|b\right| \\ Now,{\left(ab\right)}^{\left[\left|a\right|,\left|b\right|\right]}becomes: \end{gathered}$$

$\begin{array}{c}{\left(ab\right)}^{\left[\left|a\right|,\left|b\right|\right]}=a^{\left[\left|a\right|,\left|b\right|\right]}{b}^{\left[\left|a\right|,\left|b\right|\right]}\\ ={\left(a^{\left|a\right|}\right)}^{\frac{\left[\left|a\right|,\left|b\right|\right]}{\left|a\right|}}{\left(b^{\left|b\right|}\right)}^{\frac{\left[\left|a\right|,\left|b\right|\right]}{\left|b\right|}}\\ =e^{\frac{\left[\left|a\right|,\left|b\right|\right]}{\left|a\right|}}{e}^{{}^{\frac{\left[\left|a\right|,\left|b\right|\right]}{\left|b\right|}}}\\ =e\end{array}$

Therefore, $\left|ab\right||\left[\left|a\right|,\left|b\right|\right]$ divides k because it is a common multiple of$\left|a\right|and\left|b\right|$.

To show H is closed under inverses

By using exercise 7.2.28: If $a\in G,then\left|a\right|=\left|a^{-1}\right|$

As a result, we have,

$\left|a^{-1}\right|=\left|a\right|$

Hence H, is closed under inverses.

Therefore, H is closed under group operations and inverses.

By Theorem 7.11: A non-empty subset H of group is a subgroup of G, provided that:

$$\begin{gathered} 1.Ifa,b\in H,thenab\in H \\ 2.Ifa\in H,then{a}^{-1}\in H \end{gathered}$$

So, we can say that, H is a subgroup of G.