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

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 mod 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 · 4 = 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 mod 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

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

But since $|P_5|$ and $|P_3|$ are relatively prime, $P_5$ ∩ $P_3$ = e, and hence $P_5P_3$ must have order 15.

Why do we need to use the second noether theorem? Why can 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 taht $P_5P_3$ is isomorphic to $Z_{15}$? Also, how can we conclude that $P_5P_3$ normalizes $P_5$?

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For your first query, $P_{5}$ is a normal subgroup of $N_{G}(P_{5})$ and already has order $5.$ For your second query, you could compute the order as you suggest. For your third query, I am not sure what Corollary 5.3.17 says, but in any case, using Sylow's Theorem within the group $P_{3}P_{5}$, that last group can only have one Sylow $5$-subgroup, which is normal. You can then immediately derive a contradiction in any case, without worrying about the isomorphism type of $P_{5}P_{3}.$ –  Geoff Robinson Apr 5 '13 at 20:31

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 $\,P_3P_5\,$ )

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For your second answer...instead of using the Noether theorem, can I just say that...since |$P_3$|=3 and $P_3$ contains an element of order 3 and since |$P_5$|=5 and $P_5$ contains an element of order 5, they must be disjoint? Because if one element of $P_3$ other than the identity is contained in $P_5$, then the other element would also need to be contained in $P_5$. So we wouldn't be able to have an element of order 5. Is that correct or am I missing something? –  user58289 Apr 5 '13 at 20:34
Well @Artus, that's basically the reason why they deduce $\,P_3\cap P_5=\{e\}\,$ ...yes. –  DonAntonio Apr 5 '13 at 20:36
A Nice Survey and enough. –  Babak S. Apr 5 '13 at 20:43

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 $n_5=1.$

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