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


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 $n_3=1$.

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    $\begingroup$ Why does $y^{-1}x$ have order $5$? $\endgroup$ – David A. Craven Jul 28 '20 at 17:17
  • $\begingroup$ Oh... you are right. I can't guarantee its order is $5$ since I think I can't find apropiate $x$ and $y$ so that they conmute... By the way, thank you very much for your answer. $\endgroup$ – seferpd Jul 28 '20 at 18:00
  • $\begingroup$ I just edited the question to add the group-theory tag. I think group questions get answered quicker with that tag than not, but I might be wrong about that. $\endgroup$ – David A. Craven Jul 28 '20 at 18:01

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 $15$.

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    $\begingroup$ I've just added proof 6. I'm currently waiting for my computer to produce the ad-matrix of a nilpotent element of a Lie algebra, so I might come up with another proof to pass the time. $\endgroup$ – David A. Craven Jul 28 '20 at 20:38

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