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

Question

(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.

Answer 1

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.)

Answer 2

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 $n_p \cdot (p-1)$.

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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.

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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 $G$.