# Let $G$ be a group with order $105 = 3 \cdot 5 \cdot 7$

(a) Prove that a Sylow $$7$$-subgroup of $$G$$ is normal

(b) Prove that $$G$$ is Solvable

Can anyone please tell me if I am correct?

(a) For the sake of contradiction suppose $$G$$ dose not have a normal Sylow $$7$$-subgroup.

We first show $$G$$ has a normal Sylow $$5$$-subgroup. Then $$G$$ must have $$15$$ Sylow $$7$$-subgroups. So $$G$$ has $$15(7-1) = 90$$ elements of order $$7$$. If $$G$$ dose not have a normal Sylow $$5$$-subgroup then $$G$$ has $$21$$ Sylow $$5$$-subgroups so $$G$$ has $$21(5-1) = 84$$ elements of order $$5$$. But $$90 + 84 = 174 > 105$$. Therefore $$G$$ has a normal Sylow $$5$$-subgroup.

Let $$N$$ be the unique Sylow $$5$$-subroup, and let $$P$$ be a Sylow $$7$$-subgroup. Since $$N$$ is normal $$NP$$ is a subgroup of $$G$$. Since $$N \cap P = 1$$ we have $$|NP| = |N||P| = 35$$. So by Lagrange $$|G : NP| = 3$$ since $$3$$ is the smallest prime dividing $$|G|$$ we have that $$NP$$ is normal. So the Fratini Argument $$G = N_G(P)N$$

Finally since $$NP$$ is abelian $$NP$$ normalizes $$P$$. So $$NP \leq N_G(P)$$ Bur since $$3$$ divides $$|G|$$ and $$3$$ dose not divide $$N$$ we have $$3$$ divides $$N_G(P)$$ so $$105$$ divides $$N_G(P)$$ thus $$G = N_G(P)$$.

(b) Continuing with the notation above $$NP$$ is a normal subgroup of $$G$$ and $$G/NP$$ has order $$3$$ so is clearly abelian. Since $$NP$$ is a abelian, the trivial subgroup $$1$$ is a normal subgroup of $$NP$$ and $$NP/1$$ is abelian. Hence $$1 < NP < G$$ is our disired chain.

Also if anyone has any nice rules for proving that groups of a certain order are solvable that would be appreciated. I herd groups with order divisible by at most $$2$$ distinct primes must be solvable.

• How did you deduce that $NP$ is abelian? Once you know that then you are done. If there are fifteen Sylow $7$-subgroups $P$ then $N_G(P)=P$. But if $NP$ is abelian then clearly $N_G(P)$ has order at least $35$, a contradiction. As for (b), you proved that $NP$ was normal under the assumption that there are fiteen Sylows $7$-subgroups. Since that is false, $NP$ does not exist. Jul 10 '20 at 21:36
• NP is abelian since $|NP/P| = 5$, so $NP/P$ is abelian so $NP' \leq P$, where $NP'$ is the commutator subgroup of $NP$, and $|NP/N| = 7$, so $NP/N$ is abelian so $NP' \leq N$ . Since $N \cap P = 1$ we have $NP' = 1$ hence $NP$ is abelian. Jul 10 '20 at 22:10
• Why is $P$ normal in $NP$? It is if and only if $NP$ is abelian, so that's not a good start. (I agree that $NP$ is abelian, by the way, but it's a gap in your argument.) Jul 10 '20 at 22:12
• P is normal in NP because the number of Sylow $7$-subgroups of $NP$ must divide the order of $NP$ which is $35$ and must be congruent to $1$mod$7$, so there must be one Sylow $7$ subgroup, which is $P$, and when a Sylow subgroup is unique for a given prime then the subgroup must be normal. Jul 10 '20 at 22:18
• Well that's all you need. Both $N$ and $P$ are normal in $NP$, so $NP\cong N\times P$. Thus $P$ is centralized by $N$, so $N_G(P)>P$. Jul 10 '20 at 22:24

Here's another way that bypasses the question entirely. It uses the fact that the $$5$$ is a red herring, and just put there to make the numbers clash. Notice that by standard counting, groups of order $$15=3\cdot 5$$ and $$35=7\cdot 5$$ are cyclic, hence both have a normal (and unique) Sylow $$5$$-subgroup, and the same for the other prime $$3$$ or $$7$$.

We first claim that the Sylow $$p$$-subgroup is normal for some prime $$p$$. If not, then $$n_p$$, the number of Sylow $$p$$-subgroups, is given by $$n_3=7$$, $$n_5=21$$ and $$n_7=15$$. Standard element counting gives a contradiction.

If $$n_5=1$$ then $$G$$ has a normal Sylow $$5$$-subgroup. If $$n_3=1$$ or $$n_7=1$$ then $$Q\lhd G$$ where $$|Q|=3$$ or $$|Q|=7$$. Then $$G/Q$$ has order $$15$$ or $$35$$, and in both cases has a normal Sylow $$5$$-subgroup. Take the preimage of this to give a normal subgroup of $$G$$ of order $$35$$ or $$15$$. Again this has a normal Sylow $$5$$-subgroup, so again $$G$$ has a normal Sylow $$5$$-subgroup.

Quotient out by this. Then $$G$$ has order $$21$$, and easily has a normal Sylow $$7$$-subgroup. But again, take preimages to get a normal subgroup of order $$35$$, hence a normal Sylow $$7$$-subgroup as well.

Thus any group of order $$105$$ has a normal Sylow $$5$$-subgroup and a normal Sylow $$7$$-subgroup. Since the quotient, of order $$3$$, cannot act in a non-trivial way on a group of order $$5$$ (but can on a group of order $$7$$) one obtains that $$G$$ is the direct product of $$\mathbb{Z}_5$$ and a group of order $$21$$. (There are two such groups.)

• In the $3^{rd}$ paragraph I understand how if you got that if you have a normal Sylow $3$ or Sylow $7$ subgroup then you have a normal subgroup of order $35$ or $15$ which mus have a normal subgroup of order $5$, but since a normal subgroup of a normal subgroup is not necessarily normal in he whole group I don't understand how you get a normal subgroup of order $5$ this way? Jul 10 '20 at 23:10
• This is true in general, but for Sylow $p$-subgroups it is true that if they are normal in a normal subgroup then they are normal in the whole group. This is because a Sylow $p$-subgroup is unique if and only if it's normal. Being normal is something that doesn't extend transitively (what you need is characteristic for that, being invariant under all automorphisms) but being unique does. Here is an easy way to prove it: all Sylow $p$-subgroups are conjugate. So if one Sylow $p$-subgroup lies inside a normal subgroup $N$ then all of them do. So if there's one in $N$ then there's one in $G$. Jul 11 '20 at 12:38

Sylow's theorems tells us about the number $$n_p$$ of Sylow p-subgroups:

• each Sylow subgroup $$P$$ has order $$|P| = p^r$$ where $$p^r | |G|$$.
• $$n_p \equiv 1 \pmod p$$
• $$n_p | m$$ where $$m = |G|/p^r$$.
• $$n_1 = 1$$ iff $$P$$ in a normal subgroup of $$G$$.

In the case that $$r=1$$ we can say that since the Sylow p-subgroups $$P$$ are cyclic groups they will have trivial intersection. This lets us count how many elements they contribute to the group:

• The number of elements of order $$p$$ are $$n_p \cdot (p-1)$$.

For $$|G| = 3 \cdot 5 \cdot 7$$ we deduce some possibilities:

• $$n_3 = 1\text{ or }7$$

• $$n_5 = 1\text{ or }21$$

• $$n_7 = 1\text{ or }15$$

• (A) Suppose $$n_3 = 7$$ then there would be $$14$$ elements of order 3 in the group.

• (B) Suppose $$n_5 = 21$$ then there would be $$84$$ elements of order 5 in the group.

• (C) Suppose $$n_7 = 15$$ then there would be $$90$$ elements of order 7 in the group.

Clearly (B) and (C) cannot both be true, $$84 + 90 > |G| = 104$$.

Now suppose for contradiction that $$n_7 = 15$$.

• If $$n_3 = 7$$ then $$90 + 14 = 104$$ uses up all the elements of the group without leaving room for the identity or order 5 elements. Impossible.
• If $$n_3 = 1$$ then $$104 - (90 + 2 + 1) = 11$$ means there must be 11 elements of order 5 in the group, but the number of elements of order 5 must be 4 or 21. impossible.

This proves that $$n_7 = 1$$ so we have a unique normal Sylow 7-subgroup.

Regarding solvability: Let $$P$$ be the Sylow 7-subgroup. Since it's normal you can take the quotient $$|G/P| = 15$$, this is a cyclic group (because $$15$$ is relatively prime to $$\phi(15)$$) therefore abelian. This gives you a normal series for $$G$$.

• If $n_3=1$ then there are $11$ elements of order not $3$ or $7$. That includes $5$, but what about some other element order, e.g., $15$? I know this cannot occur for various reasons, but it needs to be shown. Jul 10 '20 at 22:20
• @DavidCraven Thank you for the correction, I don't fully understand what I did wrong? Was it assuming that all elements of the group except the identity lie within Sylow subgroups?
– user581023
Jul 11 '20 at 8:00
• Yes. You can normally not get exactly the right number of elements just by looking at Sylows, and indeed will not in this case. For all groups such as this there is exactly one Sylow $5$-subgroup and exactly one Sylow $7$-subgroup, so there must be lots of elements left over. Jul 11 '20 at 12:36