# Elementary Group Theory, Sylow Theorems

So it's time for me to post another elementary question. I've been stuck on this exercise for quite some time now, and really can't find a satisfactory solution for this. The section on applications of Sylow theorems in my book is kind of concise, and maybe there is some trivial fact I'm missing.

If $G$ is a group of order 60 that has a normal Sylow 3-subgroup, prove that $G$ also has a normal Sylow 5-subgroup.

Attempts: I'll denote the set of Sylow $p$-subgroups by $S_p$ and the given normal Sylow 3-subgroup by $N$. By Sylow Theorems, we know that $|S_5| = 1$ or $6$. In the first case, Sylow 5-subgroup must be normal, so I've tried to assume that $|S_5| = 6$ and somehow derive a contradiction. Since $N$ is normal in $G$, quotient group $G/N$ exists, and must have order 20. Function $f: G \rightarrow G/N$ is then our homomorphism between $G$ and it's quotient group. By inspection and ordinary counting arguments we can see that $|S_5| = 1$ in $G/N$, so there are only four elements (cosets) of order 5 in $G/N$. Assuming that $|S_5| = 6$, then we have 24 elements of order 5 in $G$. Then $f$ should map these to a coset which s order is dividing the order of 5, but that is only 1 or 5. Because index of $N$ in $G$ is 20, we get that there are 3 elements in each coset. Then we would have space for at most 15 elements in the cosets of order 1 or 5 in $G/N$, yet by assumption $G$ has 24 elements of order 5.

Am I on the right way? Is there a trivial way to prove this? I've had no bigger problems with other exercises surrounding this one, so I guess there might be a trivial way of proving this.

Attempt 2: If $N$ is the normal Sylow 3-subgroup, then $G/N$ exists and we established there exists a normal Sylow 5-subgroup $T$ of $G/N$. By Correspondence theorem $T$ is on the form $H/N$, where $H$ is a subgroup of $G$ containing $N$. Since $|H| = |H/N||N|$, I deduce $|H|=15$. Since $H/N$ is normal, quotient $(G/N)/(H/N)$ exists and is by third isomorphism theorem isomorphic to $G/H$. So if $G/H$ exists, then $H$ must be normal of $G$. We see, since $|H| = 15$, that it contains a single normal Sylow 5-subgroup. Now since the Sylow 5-subgroup I'm looking for is contained in the normal subgroup $H$, the number of Sylow 5-subgroups of $H$ must be same as in $G$ because of the Second Sylow Theorem and normality of $H$. I this the correct solution?

edit: Actually I think I've made a mistake in the italic part. $(G/N)/(H/N)$ certainly exists, but does not need to be isomorphic to $G/H$ (?). The theorem I've used assumes normality of H, so I end up in circular logic.

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You second attempt is correct and quite elegant. Note that the correspondence theorem says, among other things, that normal subgroups of the quotient correspond to normal subgroups of the big group. That's why the third isomorphism theorem always makes sense. The rest of your argument works: since $H$ is normal, $G$ permutes its Sylow 5-subgroups. But it contains only one Sylow 5-subgroup, so this Sylow 5-subgroup must also be normal in $G$. –  Alex B. Aug 26 '11 at 10:52

I think there is a problem with your solution in that it seems to assume that elements of order $n$ in $G/N$ always lift to elements of order $n$ in $G$. However, that is not true in general. What you do have is that if $gN$ has order $n$ in $G/N$, then $g$ has order divisible by $n$ and dividing $n|N|$.

Here is a suggestion for an alternative approach:

1. First show that elements of order 3 commute with elements of order 5. Indeed, if $g$ has order 3 and $h$ has order 5, then, $\langle g\rangle$ being normal, $hgh^{-1}$ is either $g$ or $g^2=g^{-1}$. If it's the latter, then what is $h^5gh^{-5}$?... Contradiction (exercise), so $gh = hg$.

2. Now, what does that tell us about the order of $gh$?

3. As you have observed, the number of Sylow 5-subgroups seems to decrease as you pass from $G$ to $G/N$, i.e. two such groups must become equal "modulo $N=\langle g\rangle$". Use Step 2. to derive a contradiction (what does it mean for two cyclic groups to become equal in the quotient?).

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thank you for your tips. I guess I will employ the Feynman problem solving algorithm and try to go further on this. Also, please excuse my beginner ignorance (and lack of instructor), but if an element $a \in G$ has order 5, does it not need to be mapped to a coset of order dividing 5 in the quotient group $G/N$, since 3 and 5-subgroups are disjoint short of the identity? –  Malman Aug 25 '11 at 19:05
@Barre Yes, that's correct, and has nothing to do with 3 and 5 subgroups. In general, if $\phi:G\rightarrow G'$ is any group homomorphism, then an element of order $n$ in $G$ is mapped to an element of order dividing $n$ in $G'$ (exercise). But if I understood your solution correctly, then you wanted to go the other way, lifting elements from $G/N$ to $G$ and saying something about the order of the lifts. Maybe I misunderstood what you were trying to do? –  Alex B. Aug 26 '11 at 5:34
I've been following up on your answer and it lead me to a slightly different 'solution'. I've added it to my original post, could you comment on that? Also, thank you for your time as always :) –  Malman Aug 26 '11 at 10:19

An alternative approach to what Alex was suggesting, is to take a look at the product $H := SN$ of $5$-Sylow subgroup $S$ of $G$ and the normal $3$-Sylow subgroup $N$ of $G$. As $N$ is normal, $H$ is a subgroup of $G$.

How many $5$-Sylow subgroups does $H$ have? What does this tell you about the normalizer of $S$ in $H$ rsp. in $G$?

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I've updated my post with a new attempt similar, but still different, to the one suggested by you. Anyway, since $S$ and $N$ are disjoint, $H = SN$ would have order 15, and containing $S$. Normalizer of $S$ is a subgroup of $H$ containing $S$, so I have options for order of the normalizer 5 or 15 (entire $H$). If it's the latter, then $S$ is normal in $H$. I'll work further on this :) –  Malman Aug 26 '11 at 10:58
@Barre: As Alex already commented, your 2nd attempt is elegant. In my opinion it is nicer than what I suggested. As you seemed to struggle with working $\bmod N$ and then lifting the result back to $G$ (which you now mastered!), I just wanted to open another road to a proof. –  Someone Aug 26 '11 at 12:06
@Barre: In your comment you try to conclude from the order of the normalizer to normality of $S$ in $H$. My questions are in the other order ;-). –  Someone Aug 26 '11 at 12:16