# Proving a group of order $35^3$ is solvable

Is my thinking correct when asked to show that a group $$G$$ of order $$35^3$$ is solvable, I first show that by the sylow theorems there exists a sylow $$p$$-subgroup of order $$5^3$$ and another unique sylow $$p$$-subgroup of order $$7^3$$. Then since these two unique sylow $$p$$-subgroups compose the group and are solvable then the entire group is solvable.

What are the techniques of showing a group is solvable with the sylow theorems?

I'm having trouble proving this one and another where the group is of order $$80$$.

Thank you for the help.

• I guess you cannot use that every group with odd order is solvable and neither Burnside's theorem stating that a group is solvable if the order has at most two distinct prime factors. Mar 27 at 20:10
• Your argument for 35^3 is correct, if you're precise about what "two subgroups compose the group" means and why that implies the whole group is solvable. For 80, you can show the 5-Sylow is unique, then use the fact for if a normal subgroup and the quotient by it are both solvable then so too is the whole group. Mar 27 at 20:15
• @runway44 In the case $\vert G \vert = 80$, it's not obvious to me that the $5$-Sylow subgroup is unique, but if it's not, then it is clear to me that the $2$-Sylow subgroup has to be, and that's good enough to get the result as long as you know that all $p$-groups are solvable. Mar 27 at 21:16
• @RobertShore Sylow theorems ensure the number $n_5$ of $5$-Sylows divides $2^4$ and is $1$ mod $5$. The only solution to that is $n_5=1$. On the other hand, they only ensure the number $n_2$ of $2$-Sylows divides $5$ and is $1$ mod $2$, which by itself is consistent with $n_2=5$ so it does not automatically imply the $2$-Sylow is normal. Mar 27 at 21:19
• @runway44 Why can't $n_5=16$? Mar 28 at 9:05

If $$n_5=1$$, then let $$P$$ be the unique $$5$$-Sylow subgroup. Both $$P$$ and $$G/P$$ are $$p$$-groups, so $$G$$ is solvable.

If $$n_5=16$$, then each $$5$$-Sylow subgroup of $$G$$ has trivial intersection with each other $$5$$-Sylow subgroup so there are $$4 \cdot 16 = 64$$ elements of order $$5$$ in $$G$$. Since a $$2$$-Sylow subgroup must have $$15$$ non-identity elements, $$G$$ must have a unique $$2$$-Sylow subgroup, $$Q$$. Both $$Q$$ and $$G/Q$$ are $$p$$-groups, so $$G$$ is solvable.

All of the Sylow subgroups of a group are unique if and only if it is the direct product of those subgroups.

But Sylow subgroups are $$p$$-groups; which are solvable.

Finally the direct product of solvable groups is again solvable.

Note that for a group of order $$80$$ it suffices to get one normal Sylow, since then it and the quotient group by it are $$p$$-groups.

• Which theorem of Burnside are you talking about? The fact that $p$-groups are solvable is easy to prove, and not usually attributed to Burnside - Cauchy is more likely. Mar 27 at 22:00
• @DerekHolt I was referring to the one that says groups of order $p^aq^b$ are solvable. I figured it was overkill.
– user403337
Mar 27 at 22:05
• Of course that theorem solves the original problem, all by itself. @DerekHolt
– user403337
Mar 27 at 22:18
• How do we show a group of order $80$ is a direct product of its Sylows? Mar 28 at 0:48
• I suppose it may not be. @runway44 for instance we could do a semi-direct product $\mathbb Z_5\rtimes\mathbb Z_{16}$.
– user403337
Mar 28 at 2:04