Given a Group, Prove Subgroup

[Mathyes]

Would someone please help check my solution here?

The problem: Given a group G with a subgroup \(\displaystyle H \le G\) (H is a subgroup of G). Suppose we fix an element \(\displaystyle {g_0} \in G\). Show that \(\displaystyle {g_0}Hg_0^{ - 1} = \{ ({g_0}hg_0^{ - 1})|h \in H\} \) is a subgroup of G.

Solution:
To prove \(\displaystyle {g_0}Hg_0^{ - 1}\) is a subgroup of G, I need to show three things. They are \(\displaystyle {g_0}Hg_0^{ - 1}\) is closed under the operation of G; the identity \(\displaystyle {e_G} \in {g_0}Hg_0^{ - 1}\), and the inverse. I have done the first two already; however, the inverse is the one I am not sure of.
Here is my showing of the inverse:
I need to show that

\(\displaystyle \forall k \in {g_0}Hg_0^{ - 1},{k^{ - 1}} \in {g_0}Hg_0^{ - 1}\) that is \(\displaystyle {k^{ - 1}} = {g_0}{h^{ - 1}}g_0^{ - 1}\)
so
\(\displaystyle k = {g_0}hg_0^{ - 1}\) \(\displaystyle h \in H\)
\(\displaystyle {k^{ - 1}} = {({g_0}hg_0^{ - 1})^{ - 1}}\)
\(\displaystyle {k^{ - 1}} = {g_0}{h^{ - 1}}g_0^{ - 1}\) (\(\displaystyle {h^{ - 1}} \in H\))

Therefore, \(\displaystyle \forall k \in {g_0}Hg_0^{ - 1},{k^{ - 1}} \in {g_0}Hg_0^{ - 1}\)

 

[pka]

Would someone please help check my solution here?
The problem: Given a group G with a subgroup \(\displaystyle H \le G\) (H is a subgroup of G). Suppose we fix an element \(\displaystyle {g_0} \in G\). Show that \(\displaystyle {g_0}Hg_0^{ - 1} = \{ ({g_0}hg_0^{ - 1})|h \in H\} \) is a subgroup of G.
Solution:
To prove \(\displaystyle {g_0}Hg_0^{ - 1}\) is a subgroup of G, I need to show three things. They are \(\displaystyle {g_0}Hg_0^{ - 1}\) is closed under the operation of G; the identity \(\displaystyle {e_G} \in {g_0}Hg_0^{ - 1}\), and the inverse. I have done the first two already; however, the inverse is the one I am not sure of.
Here is my showing of the inverse:
I need to show that
\(\displaystyle \forall k \in {g_0}Hg_0^{ - 1},{k^{ - 1}} \in {g_0}Hg_0^{ - 1}\) that is \(\displaystyle {k^{ - 1}} = {g_0}{h^{ - 1}}g_0^{ - 1}\)
so
\(\displaystyle k = {g_0}hg_0^{ - 1}\) \(\displaystyle h \in H\)
\(\displaystyle {k^{ - 1}} = {({g_0}hg_0^{ - 1})^{ - 1}}\)
\(\displaystyle {k^{ - 1}} = {g_0}{h^{ - 1}}g_0^{ - 1}\) (\(\displaystyle {h^{ - 1}} \in H\))
Therefore, \(\displaystyle \forall k \in {g_0}Hg_0^{ - 1},{k^{ - 1}} \in {g_0}Hg_0^{ - 1}\)

Yes that works, even if somewhat messy.

Here is a much easier way:
Theorem: A set \(\displaystyle \mathcal{H}\) is a subset of a group \(\displaystyle \mathcal{G}\) is a subgroup iff \(\displaystyle \{x,y\}\subset \mathcal{H}\) implies \(\displaystyle xy^{-1}\in\mathcal{H}\).

 

[Mathyes]

Thank you Pka