# A detail in a proof about isomorphism classes of groups of order 21

1. Mar 20, 2014

### Whovian

1. The problem statement, all variables and given/known data

While reading through my textbook on abstract algebra while studying for a test, I ran across the following statement:

There are two isomorphism classes of groups of order 21: the class of $C_{21}$, and the class of a group $G$ generated by two elements $x$ and $y$ that satisfy the relations $x^7=1$, $y^3=1$, $yx=x^2y$

along with a proof of this statement. The proof looks fairly reasonable, only there's a bit I don't quite understand:

"The Third Sylow Theorem shows us that the Sylow 7-subgroup $K$ [existence and uniqueness guaranteed by the Third Sylow Theorem] must be normal ..."

Specifically, why must $K\trianglelefteq G$?

2. Relevant equations

Third Sylow Theorem: If $G$ is a group of order $n<\aleph_0$, where $n=p^e\cdot m$ for some $e>0$ and $\lnot\left(p|m\right)$, for some prime $p$, and $s$ is the number of Sylow $p$-subgroups of $G$, then $s|m$ and $s\equiv1\pmod p$.

3. The attempt at a solution

Fairly nonexistent; I can't think of any reason this subgroup must be normal.

(If this post would be more appropriate in the linear & abstract algebra section, feel free to move it there.)

2. Mar 20, 2014

### Dick

Use what they are saying about $s$ to show there must be only one 7-subgroup. So?

3. Mar 20, 2014

### Whovian

There must be a 7-subgroup; I just don't see why it must be a normal subgroup.

4. Mar 20, 2014

### Dick

Use the definition of normal. There is only ONE 7-subgroup.

5. Mar 20, 2014

### Whovian

I'm used to normal being defined as "closed under conjugation by elements of G." Is there a different or equivalent definition I'm missing or something else obvious which makes clear that only one 7-subgroup existing means said subgroup is normal? (Knowing me, most likely.)

6. Mar 20, 2014

### jbunniii

If you conjugate a subgroup, the result is another subgroup of the same size. So if there's only one subgroup of order 7...