Show H is equal or less than N(H).

[mcwang719]

Let G be a group and let H be a subgroup of G.

N(H)={x belongs G / xHx^-1 = H}
a)Show that H is equal or less than N(H).
b) Prove that N(H) is a subgroup of G.

To be honest I'm completely lost on this problem, I've read the chapter, but still think I'm missing something. If anyone could help me get started that would be great. thanks

 

[pka]

Here is a theorem.
A subset H of a group G is a subgroup of G if and only if when a & b are in H then ab<SUP>-1</SUP> is also in H.

If a & b are in N(H) then \(\displaystyle aHa^{ - 1} = H\quad \& \quad bHb^{ - 1} = H.\)

\(\displaystyle \L \begin{array}{c}
\left( {ab^{ - 1} } \right)H\left( {ab^{ - 1} } \right)^{ - 1} \\
\left( {ab^{ - 1} } \right)H\left( {ba^{ - 1} } \right) \\
a\left( {bHb^{ - 1} } \right)a^{ - 1} \\
a\left( H \right)a^{ - 1} \\
H \\
\end{array}\)

 

[mcwang719]

Thanks Pka! That helped out a lot. Can you give me some pointers to show H is equal or less than N(H).

 

[daon]

Thanks Pka! That helped out a lot. Can you give me some pointers to show H is equal or less than N(H).

First part is easy, just show H \(\displaystyle \subseteq\) N(H). This is since you are already given H is a subgroup of G. If H and N(H) are subgroups and you want to show H \(\displaystyle \le\) N(H), then all you need to do is show that all of H is inside N(H). If they introduced H as only a subset of G and not a subgroup of g, we couldn't do this. So anyway, take an arbitrary element of H and show that it is in N(H).

Also that sign you're referring to means "is a subgroup of" not "less-than-or-equal-to". Also, read the chapter again. I know algebra is a tough subject (I just took it). Take it from me that just reading the chapter once won't cut it. Try to do proofs in the book before you read them too, it helps.