Question

Let be a subgroup of a groupand a subgroup of a group. Prove that is a subgroup GX H.

Short answer

Answer

It is proved that is a subgroup of GX H.

Step-by-step solution

Step by Step Solution: Step 1: Show that G1 X H1 is closed under the group operation

We know that is a subset of G XH.

Therefore, by using Theorem 7.11, we will check that

1. If x,y $\in$H then xy $\in$H
2. If x $\in$H, then x^-1$\in$H

Let us check the first one:

1. Let$x=\left(g_1,h_1\right)$and be two arbitrary elements of G₁XH₁, i.e.,

Then,

$x\cdot y=\left(g_1{g}_2,h_1{h}_2\right)\in G_1\times H_1\dots \left(\begin{array}{l}\because g_1{g}_2\in G_1\text{\hspace{0.17em}\hspace{0.17em}since\hspace{0.17em}}{G}_1\text{\hspace{0.17em}\hspace{0.17em}is\hspace{0.17em}\hspace{0.17em}a\hspace{0.17em}\hspace{0.17em}subgroup\hspace{0.17em}\hspace{0.17em}of\hspace{0.17em}\hspace{0.17em}}G\\ \text{and\hspace{0.17em}\hspace{0.17em}}{h}_1{h}_2\in H_1\text{\hspace{0.17em}since\hspace{0.17em}\hspace{0.17em}}{H}_1\text{\hspace{0.17em}\hspace{0.17em}is\hspace{0.17em}\hspace{0.17em}a\hspace{0.17em}\hspace{0.17em}subgroup\hspace{0.17em}\hspace{0.17em}of\hspace{0.17em}\hspace{0.17em}}H\end{array}\right)$

Hence,G₁XH₁ is closed under group operation.

Show that G1 X H1  is closed under inverse

Now, $\begin{array}{l}x=\left(g_1,h_1\right)\in G_1\times H_1\\ \therefore x^{-1}=\left(g_1^{-1},h_1^{-1}\right)\in G_1\times H_1\dots \left(\begin{array}{l}\because g_1^{-1}\in G_1\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}since\hspace{0.17em}\hspace{0.17em}}{G}_1\text{\hspace{0.17em}\hspace{0.17em}is\hspace{0.17em}\hspace{0.17em}a\hspace{0.17em}\hspace{0.17em}subgroup\hspace{0.17em}\hspace{0.17em}of\hspace{0.17em}\hspace{0.17em}}G\\ \text{and\hspace{0.17em}\hspace{0.17em}}{h}_1^{-1}\in H_1\text{\hspace{0.17em}\hspace{0.17em}since\hspace{0.17em}\hspace{0.17em}}{H}_1\text{\hspace{0.17em}\hspace{0.17em}is\hspace{0.17em}\hspace{0.17em}a\hspace{0.17em}\hspace{0.17em}subgroup\hspace{0.17em}\hspace{0.17em}of\hspace{0.17em}\hspace{0.17em}}H\end{array}\right)\\ \end{array}$

Hence, G₁XH₁ is closed under inverse.

Therefore,G₁XH₁ is closed under group operation and inverse.

Hence, G₁XH₁ is a subgroup of G₁XH.