# 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 $$n_3=1$$.

• Why does $y^{-1}x$ have order $5$? – David A. Craven Jul 28 '20 at 17:17
• 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. – seferpd Jul 28 '20 at 18:00
• 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. – David A. Craven Jul 28 '20 at 18:01

## 1 Answer

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 $$15$$.

• 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. – David A. Craven Jul 28 '20 at 20:38