A group of order $30$ has a normal $5$-Sylow subgroup.

Question

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.

Answer 1

Short Answers:

1.- Because $\,P_5\le N_G(P_5)\,$ and $\,|N_G(P_5)|=|P_5|\,$

2.- The "2nd Noether Theorem" seems to be what others (like me) call one (the second or the third, usually) of the isomorphism theorems, and the formula you want to use is precisely the order of the group

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

otherwise you wouldn't be able to deduce $\,|P_3P_5|=15\,$ ...

3.- The group $\,\Bbb Z_{15}-\,$ the cyclic group of order $\,15\,$ is abelian , so it trivially normalizes its own subgroups...(this means $\,P_5\,$ is normal in $\,P_3P_5\,$ )

Answer 2

$o(G) = 30 = 2*3*5$

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

Answer 3

This very question showed up on my test today. And this is the proof I wrote:

For the sake of contradiction, assume that a group $G$ of order $30$ does not have unique Sylow $5$-subgroup.

By Sylow theory and order counting argument, we can conclude that $G$ has six Sylow $5$-subgroups (say, $H_1,\ldots,H_6$) and one Sylow $3$-subgroup (say, $K$).

Now, $K\lhd G$ so we have subgroups $KH_1,\ldots,KH_6$ (each of order $15$) and they're all distinct (because a group of order $15$ has a unique Sylow $5$-subgroup and for each $KH_i$, its unqiue Sylow $5$-subgroup is $H_i$).

Now, consider the group $G/K$. We see that $KH_1/K$, $\ldots$, $KH_6/K$ are distinct subgroups of $G/K$ and they're each of order $5$. This is clearly a contradiction since a group of order $10$ must have unique Sylow $5$-subgroup. $\blacksquare$

Answer 4

A group of order 30 has a normal 5-Sylow subgroup.

$|G|=2.3.5$

The divisors of $|G|$ are $1,2,3,5,6,10,15,30.$

$5|n_5-1\implies n_5=1,6\\3|n_3-1\implies n_3=1,10\\2|n_2-1\implies n_2=1,3,5,15.$

If possible let $n_5=6.$ Then $G$ has at least $5.6-5=25$ elements of order $5.$ Consequently since $|G|=30,n_3=n_2=1.$ Let $H_2,H_3$ be normal Sylow $2$ and Sylow $3$ subgroups of $G.$ Since $H_1H_2\le G$ and $H_2H_3=H_1\times H_2\simeq H_2\oplus H_3, $ $H_2H_3$ has $6$ elements none of which is of order $5.$ Thus $G$ has at least $25+6=31$ elements! So $n_5=1.$