Question

If N is a normal subgroup of a group G and T is a subgroup of$\mathit{G}\mathbf{/}\mathit{N}$ , show that $\mathit{H}\mathbf{=}\mathbf{\left\{}\mathbf{a}\mathbf{\in }\mathbf{G}\mathbf{|}\mathbf{N}\mathbf{a}\mathbf{\in }\mathbf{T}\mathbf{\right\}}$is a subgroup of G.

Short answer

It is proved that, $H=\left\{a\in G|Na\in T\right\}$is a subgroup of G.

Step-by-step solution

Important Theorems

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${\mathit{a}}^{\mathbf{-}\mathbf{1}}\mathbf{\in }\mathit{H}$.

Theorem 8.11

The following conditions on a subgroup $\mathit{N}$of a group $\mathit{G}$are equivalent:

1. Nis a normal subgroup of G.
2. $a^{-1}Na\subseteq N$for every $\mathit{a}\mathbf{\in }\mathbf{\text{\hspace{0.17em}\hspace{0.17em}}}\mathit{G}$, where${\mathit{a}}^{\mathbf{-}\mathbf{1}}\mathit{N}\mathit{a}\mathbf{\subseteq }\mathit{N}$$\mathbf{\left\{}{\mathbf{a}}^{\mathbf{-}\mathbf{1}}\mathbf{N}\mathbf{a}\mathbf{|}\mathbf{n}\mathbf{\in }\mathbf{N}\mathbf{\right\}}$.
3. $\mathit{a}\mathit{N}{\mathit{a}}^{\mathbf{-}\mathbf{1}}\mathbf{\subseteq }\mathit{N}$for every$\mathit{a}\mathbf{\in }\mathbf{\text{\hspace{0.17em}\hspace{0.17em}}}\mathit{G}$, where localid="1659513820752" $\mathit{a}\mathit{N}{\mathit{a}}^{\mathbf{-}\mathbf{1}}\mathbf{\subseteq }\mathit{N}\mathbf{,}$$\mathbf{\left\{}\mathbf{a}\mathbf{N}{\mathbf{a}}^{\mathbf{-}\mathbf{1}}\mathbf{|}\mathbf{n}\mathbf{\in }\mathbf{N}\mathbf{\right\}}$.
4. ${\mathit{a}}^{\mathbf{-}\mathbf{1}}\mathit{N}\mathit{a}\mathbf{\subseteq }\mathit{N}$for every$\mathit{a}\mathbf{\in }\mathbf{\text{\hspace{0.17em}\hspace{0.17em}}}\mathit{G}$.
5. $\mathit{a}\mathit{N}{\mathit{a}}^{\mathbf{-}\mathbf{1}}\mathbf{\subseteq }\mathit{N}\mathbf{,}$for every$\mathit{a}\mathbf{\in }\mathbf{\text{\hspace{0.17em}\hspace{0.17em}}}\mathit{G}$.

Given that N is a normal subgroup of group G and T is a subgroup of G/N.

Proving thatH= {a∈ G |  Na∈ T} is a subgroup of G

Since N is a normal subgroup of group G, for any$a,b\in G$,$aN{a}^{-1}\subseteq N$.

Which implies$aN{a}^{-1}\in N$.

Now, for any$a\in G$, we have:

$\begin{array}{c}aN{a}^{-1}\in N\\ aN{a}^{-1}a\in Na\\ aN\in T\end{array}$

Since$aN={\left(Na\right)}^{-1}$, ${\left(Na\right)}^{-1}\in T$.

Hence, from above theorem, it can be concluded that$H=\left\{a\in G|Na\in T\right\}$ is a subgroup of G.