TITLE: A group of order $30$ has a normal $5$-Sylow subgroup.
QUESTION [10 upvotes]: There are several things that confuse me about this proof, so I was wondering if anybody could clarify them for me.

Lemma Let $G$ be a group of order $30$. Then the $5$-Sylow subgroup of $G$ is
normal.


Proof We argue by contradiction. Let $P_5$ be a $5$-Sylow subgroup of $G$. Then the
number of conjugates of $P_5$ is congruent to $1 \bmod 5$ and divides $6$. Thus, there must be six conjugates of $P_5$. Since the number of conjugates is the index of the normalizer, we see that $N_G(P_5) = P_5$.

Why does the fact that the order of $N_G(P_5)$ is 5 mean that it is equal to $P_5$?

Since the $5$-Sylow subgroups of $G$ have order $5$, any two of them intersect in the
identity element only. Thus, there are $6\cdot4 = 24$ elements in $G$ of order $5$. This leaves $6$ elements whose order does not equal $5$. We claim now that the $3$-Sylow subgroup, $P_3$, must be normal in $G$.


The number of conjugates of $P_3$ is congruent to $1 \bmod 3$ and divides $10$. Thus, if
$P_3$ is not normal, it must have $10$ conjugates. But this would give $20$ elements of order $3$ when there cannot be more than $6$ elements of order unequal to $5$ so that $P_3$ must indeed be normal.


But then $P_5$ normalizes $P_3$, and hence $P_5P_3$ is a subgroup of $G$. Moreover, the Second Noether Theorem gives


$(P_5P_3)/P_3 \cong P_5/(P_5 \cap P_3)$.


But since $|P_5|$ and $|P_3|$ are relatively prime, $P_5 \cap P_3 = 1$, and hence $P_5P_3$ must have order $15$.

Why do we need to use the second Noether theorem? Why can't we just use the formula $\frac{|P_5||P_3|}{|P_3 \cap P_5|}$ to compute the order?

Thus, $P_5P_3 \cong Z_{15}$, by Corollary $5.3.17$. But then $P_5P_3$ normalizes $P_5$, which contradicts the earlier statement that $N_G(P_5)$ has order $5$.

Why do we have to realize that $P_5P_3$ is isomorphic to $Z_{15}$? Also, how can we conclude that $P_5P_3$ normalizes $P_5$?
Thanks in advance.

REPLY [2 votes]: $o(G) = 30 = 2*3*5$
Since the number of $5$ - Sylow subgroups is of form, $N_G(5) = 5k +1 $ and $N_G(5) \Big |  o(G)$,
$\implies N_G(5) = 1$ or $N_G(5) = 6$
Similarly, $N_G(3) = 1$ or $N_G(3) = 10$.
First we will show that, $5$ - Sylow subgroup or $3$ - Sylow subgroup is normal is $G$.
If $N_G(5) = 6$ and $N_G(3) = 10$, then,
There are $(6*4 = 24)$ elements of order $5$ and  $(10*2 = 20)$ elements of order $3$.
So a total of $24+20 =44$  elements. Which is a contradiction.
Thus, $N_G(5) = 1$ or $N_G(3) = 1$ or both.
If $N_G(5) = 1 $, then we are done. If not,
We have $N_G(3) = 1$ and $N_G(5) = 6$, let $H_3$ and $H_5$ be $3$-sylow and $5$-sylow subgroups respectively. Since $N_G(3) = 1$, $H_3$ is normal.
It is easy to see that $H_3H_5$ is a subgroup in $G$ and $o(H_3 H_5) = 15$.
By the argument used earlier, $H_5$ is normal in $H_3 H_5$. (Because, the number of 5-Syllow subgroups in $H_3H_5$, i.e., $N_{H_3H_5}(5) = 1$)
$\implies$ There are only $4$ elements of order $5$ in $H_3H_5$.
Thus, the remaining $20$ elements of order $5$ lie in $G - H_3H_5$, but $\Big | G- H_3 H_5 \Big | = 15$, a contradiction.
Hence $N_G(5) = 1$
QED