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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.

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    $\begingroup$ 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. $\endgroup$
    – Peter
    Mar 27 at 20:10
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    $\begingroup$ 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. $\endgroup$
    – runway44
    Mar 27 at 20:15
  • $\begingroup$ @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. $\endgroup$ Mar 27 at 21:16
  • $\begingroup$ @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. $\endgroup$
    – runway44
    Mar 27 at 21:19
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    $\begingroup$ @runway44 Why can't $n_5=16$? $\endgroup$ Mar 28 at 9:05
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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.

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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.

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

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