Show that if $|G|=30$ then $G$ has normal $3$-Sylow and $5$-Sylow I know this have been asked. But I would like to verify my proof:
Let $n_3$ and $n_5$ be the number of $3$-Sylow and $5$-Sylow of $G$.
I can prove using Sylow's theorems that we have only two options for $n_5$. We have $n_5=1$ or $n_5=6$. Also I know that $n_5=6$ implies $n_3=1$.
So let's suppose that $n_5=6$. Let $H$ be the normal $3$-Sylow of $G$. Then
$$|G/H|=10.$$
Let $C$ be the set of elements of order $5$ in $G$. Since $n_5=6$, we have $|C|=24$. Then there must be $x,y \in C$, with $x\neq y$, such that
$$xH=yH$$
Therefore, there exists $h \in H$ such that $x=yh$. This implies that
$$y^{-1}x=h$$
But this is a contradiction since $y^{-1}x$ has order $5$ and $h$ has order $3$.
Therefore we must have $n_5=1.$
The same kind of argument allows me to prove that $n_3=1$. This time Sylow's theorems tell us that if $n_3\neq1$ then $n_3=10$ and $n_5=1$.  Also, in this case we have that $G$ has $2\cdot 10=20$ elements of order $2$.
If $K$ is the only one $5$-Sylow then
$$|G/K|=6$$
Therefore there exists two distincs $a,b\in G$ of order $2$ such that $aK=bK$. So $b^{-1}a=k$ for some $k\in K$. Which is a contradiction since $b^{-1}a$ has order $2$ and $k$ has order $5$.
Therefore $n_3=1$.
 A: Here's one more Proof.
Let $G$ act on $G$ by left multiplication. Then there is an induced homomorphism $\varphi:G\to\operatorname{Sym}(G)\simeq S_{30}$. Let $g\in G$ be an element of order $2$. Then $\varphi(g)\in S_{30}$ is a product of $15$ transpositions. In particular, $\operatorname{sgn}(\varphi(g)) = -1$. Hence the map $\operatorname{sgn}\circ\varphi: G\to\{\pm 1\}$ is surjective. The kernel $K$ of this map has order $15$ so that $K$ is cyclic. Since each Sylow $3$ and $5$ subgroups of $K$ are normal in $K$ and $K$ is normal in $G$, both Sylow subgroups are normal in $G$. Clearly both Sylow subgroups are also Sylow subgroups of $G$.
A: OK, let's give the jumbo post of all of the proofs that a group of order $30$ has normal $3$- and $5$-subgroups.
Proof 1: Element Counting
Suppose that $G$ does not have a normal Sylow $5$-subgroup. Then $G$ has six Sylow $5$-subgroups, so has $6\times(5-1)=24$ elements of order $5$. It cannot have ten Sylow $3$-subgroups, as that would contribute another twenty elements of order $3$. Thus $G$ has a normal Sylow $3$-subgroup. This gives us another two elements, plus the identity, so $27$ in total. Thus the number of Sylow $2$-subgroups is $1$ or $3$. Either way, there is an element $y$ of order $5$ that normalizes $P$ of order $2$. Any element normalizing $P$ centralizes it (what else can you do to $\{1,x\}$?) and so $x$ centralizes $y$. Thus $x\in N_G(\langle y\rangle)$, contradicting that $G$ has six Sylow $5$-subgroups (and hence is self-normalizing.
Thus $G$ has a normal Sylow $5$-subgroup. If it has ten Sylow $3$-subgroups, then this yields 25 elements so far, with just elements of order $2$ (and any other order) to go. If there are five Sylow $2$-subgroups then we get a contradiction as before (as $3$ divides the order of $N_G(P)$) so there are three Sylow $2$-subgroups. This leaves just two elements, which must have composite order, $6$, $10$ or $15$. They cannot have order $6$ or $15$, because they would then lie in the normalizer of a Sylow $3$-subgroup, which is of order $3$. Thus they have order $10$. But in a cyclic group of order $10$ there are four elements of order $10$, not two. This means that there is also a normal Sylow $3$-subgroup.
Proof 2: Use a normal $3$- or $5$-subgroup
As above, it's easy to prove that there is either a normal Sylow $3$-subgroup or a normal Sylow $5$-subgroup $P$. Thus if $Q$ is any Sylow $p$-subgroup where $p$ is the other one from $3$ and $5$ (whichever is not necessarily normal) the group $PQ$ exists. Groups of order $15$ are cyclic, and so $PQ$ is $P\times Q$, and $N_G(Q)\geq PQ>Q$. Thus both $P$ and $Q$ are normal in $G$.
Proof 3: Use Cayley's theorem
Let $G$ be any group of order $2n$, where $n$ is odd. In 1878 (maybe) Cayley proved that $G$ has a normal subgroup of index $2$. This follows by considering the regular representation of $G$ on itself, and noting that an element of order $2$ is an odd permutation. Thus our $G$ has a subgroup of order $15$, necessarily cyclic, and so both $3$- and $5$-subgroups have the other in their normalizer. Thus $n_3=n_5=1$.
Proof 4: Proofs 2+3
Use Proof 2 to obtain a subgroup of index $2$, then use Proof 3. (This bypasses Cayley's theorem.)
Proof 5: No action
Note that if $P$ and $Q$ have orders $3$ and $5$ respectively, then there is no way for $P$ to normalize $Q$ without centralizing it, and vice versa. Thus if either $P$ or $Q$ is normal, then it is centralized by $Q$ or $P$. In particular, $PQ$ centralizes both $P$ and $Q$. So we cannot have $n_3=10$ or $n_5=6$, and $n_3=n_5=1$.
Proof 6: The Sylow $2$-subgroup
This proof counts the number of Sylow $2$-subgroups $P$. Of course, as we have seen before, $N_G(P)=C_G(P)$, since $|P|=2$. We have that $n_2\in \{1,3,5,15\}$. If $n_2=15$ then there is not enough room for either $n_3=10$ or $n_5=6$, so $n_3=n_5=1$. If $n_2=5$ then $|C_G(P)|=6$. If $n_5=6$ then we have 24 elements of order $5$, and $|C_G(P)|=6$, which is all elements. But where are the other elements of order $2$? If $n_2=3$ then $|C_G(P)|=10$. Again, if $n_3=10$ then $G$ is the union of elements of order $3$ and $C_G(P)$, and we have no more involutions. Thus $c_2=1$, $G$ has a central element of order $2$. Thus $n_3\neq 10$ and $n_5\neq 6$, since both have $P$ in their centralizer. Thus $n_3=n_5=1$.
Proof 7: Conjugation on Sylow subgroups
Suppose that there are six Sylow $5$-subgroups $P$. This gives a map from $G$ to $S_6$. Since $C_G(P)=P$, this map is faithful, so $G\leq S_6$. $G$ is transitive on its Sylow $5$-subgroups, and certainly contains a $5$-cycle. Thus $G$ is sharply $2$-transitive. There are no elements of order $2$ in $A_6$ that fix at most one point, so $G\cap A_6$ has order $15$. Such groups are cyclic, so $G$ has a normal Sylow $5$-subgroup.
Similarly, if $n_3=10$, we again have that an element of order $2$ acts as a product of five $2$-cycl;es, thus is odd, and so $A_{10}\cap G$ has order $15$.
