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.
 A: 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:


*

*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 $gh = hg$.

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

*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?).
A: 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 $G$?
