Question

Question: Let $\mathit{N}$be a subgroup of a group G and let $\mathit{a}\mathbf{\in }\mathit{G}$.

(A) Prove that${\mathit{a}}^{\mathbf{-}\mathbf{1}}\mathit{N}\mathit{a}\mathbf{=}\mathbf{\left\{}{\mathit{a}}^{\mathbf{-}\mathbf{1}}\mathit{n}\mathit{a}\mathbf{\right|}\mathit{n}\mathbf{\in }\mathit{N}\mathbf{\}}$is a subgroup of .

Short answer

It has been proved that$a^{-1}Na=\left\{a^{-1}na\right|n\in N\}$is a subgroup of G.

Step-by-step solution

Step-By-Step SolutionStep 1: Show that a-1na is non-empty

It is clear that at least $a^{-1}{e}_{G}a=e_G$is contained in $a^{-1}Na$

Now,

$$\begin{gathered} \left(a^{-1}{n}_{1}a\right)\left(a^{-1}{n}_{2}a\right)=a^{-1}{n}_1\left(a{a}^{-1}\right)n_{2}a \\ =a^{-1}{n}_1{n}_{2}a \end{gathered}$$

Here, $n_1{n}_2\in N$ since N is a subgroup.

Hence,$a^{-1}Na$is closed

Show that  a-1Nais closed under inverses

Let, $a^{-1}na\in a^{-1}Na$

Let’s find out the inverse

$$\begin{gathered} {\left(a^{-1}na\right)}^{-1}=a^{-1}{n}^{-1}{\left(a^{-1}\right)}^{-1} \\ =a^{-1}{n}^{-1}a \end{gathered}$$

Since N is a subgroup $n^{-1}\in N$

so $a^{-1}{n}^{-1}a\in a^{-1}Na$

hence,$a^{-1}Na$is closed under inverses.

Conclusion

hence,$a^{-1}Na$is a subgroup of G.