Finite normal subgroup and subgroup test I was tutoring a student today and this problem came up so I need to explain to him using simple ideas but it seems trivial. The question is:
Let $N$ be a finite subgroup of $G$. Show that if $gNg^{-1}\subset N$ then $gNg^{-1} =N$. 
I told him that Since the supposition shows that $N$ is a normal subgroup, we automatically get the result by the normal subgroup test (which he said he's proven in class). He asked me what $N$ being finite had to do with anything and I didn't really have an answer for him since this is true for any order... any thoughts?
 A: Since $N$ is finite, you can show this by counting elements. Since there are $|N|$ distinct elements on both the left and the right, and you have inclusion, you get equality.
Without finiteness, this is slightly more annoying to prove. One might prove it by noting that $gNg^{-1} \subset N$ for every $g$ gives also $g^{-1}Ng \subset N$, which rearranges to $N \subset gNg^{-1}$. Clearly, if both sets contain each other, they are equal.
A: My interpretation of the problem is to show that if a finite subgroup contains one of its conjugates, then it is equal to that conjugate. The normal subgroup test tells us that if a subgroup contains all of its conjugates then it is normal and hence equal to all of its conjugates. However, it is possible for a (necessarily infinite) subgroup to properly contain one of its conjugates. Here is an example:
Define a binary operation on the set $G = \mathbb{Q} \times \mathbb{Z}$ by
$(a,m)(b,n) = (a +2^mb, m+n)$ for all $a,b \in \mathbb{Q}$ and $m,n \in \mathbb{Z}$. Show that $G$ with this operation is a group.
It is easy to see that $(a,0)^n = (na,0)$ for every $a \in \mathbb{Q}$ and $n \in \mathbb{Z}$. It follows that $\langle (a,0) \rangle = \{(na,0) \mid n \in \mathbb{Z}\}$.
Let $x = (1,0)$ and $y = (0,1)$. Compute that $yxy^{-1} = (2,0)$. Let $H = \langle x \rangle = \{(n,0)\mid n \in \mathbb{Z}\}$. Then  $yHy^{-1} = \langle yxy^{-1} \rangle = \{(2n,0)\mid n \in \mathbb{Z}\}$ is a proper subgroup of $H$.
In fact if we define $H_n = y^nHy^{-n}$ for all $n \in \mathbb{Z}$ then we have $H_{n+1}$ is a proper subgroup of $H_n$ for all $n \in \mathbb{Z}$, and all of these subgroups are conjugate.
