Let $G$ be a group of order $20$ and its class equation is given by $$1+4+5+5+5$$

My question is whether Sylow-$2$ subgroup of $G$ is normal or not.

By Sylow theorem we know that the Sylow-$2$ subgroup will be normal if and only if there is only one Sylow-$2$ subgroup inside $G$. Sylow theorem says that number Sylow-$2$ subgroup is $2k+1$ where $2k+1$ divides $5$. Only $1$ and $5$ are possible. How to decide what will be the case here.

Thank you!

  • $\begingroup$ Sylow states there are $1,3$ or $5$ Sylow-$2$ subgroups (a group of size 4 here is a Sylow-$2$ subgroup because 4 is the highest power of 2 dividing 20) $\endgroup$ – Max Freiburghaus May 23 '18 at 13:16
  • $\begingroup$ @MaxFreiburghaus Since $3$ does not divide $5$, there is only either $1$ or $5$ Sylow $5$-subgroups. $\endgroup$ – Bill Wallis May 23 '18 at 13:18
  • $\begingroup$ oh, of course. Fair point $\endgroup$ – Max Freiburghaus May 23 '18 at 13:19
  • $\begingroup$ Oops, I meant Sylow $2$-subgroups in my comment. $\endgroup$ – Bill Wallis May 23 '18 at 13:39

By Sylow, you can easily find that there is one Sylow $5$-subgroup, so the only Sylow $5$-subgroup, say $P$, is normal in your group $G$.

Now if your Sylow $2$-subgroup, say $Q$, is normal, then we must have that $G$ is the direct product of its Sylow $p$-subgroups, so that $G = P \times Q$. Now $P$ has order $5$ so it's cyclic and $P \cong C_{5}$. On the other hand, we have that $Q$ is order $4$ so either $Q \cong C_{4}$ or $Q \cong C_{2} \times C_{2}$. Thus the only order $20$ groups with a normal Sylow $2$-subgroup are $$ G \cong C_{4} \times C_{5} \cong C_{20} \quad\text{or}\quad G \cong C_{2} \times C_{2} \times C_{5} \cong C_{2} \times C_{10}. $$

The other order $20$ groups are not the direct product of the Sylow $p$-subgroups, so they can't have unique (hence normal) Sylow $p$-subgroups. I'm not sure what the class equation is, but if you can use that to determine properties of your group, you can determine whether it's one of the two above or not.

  • $\begingroup$ I think there will be no Sylow $2$- normal subgroup because if it was normal then by your logic the group $G$ will be abelian and an abelian group cannot have a class equation of the above type? $\endgroup$ – user276115 May 23 '18 at 17:36
  • $\begingroup$ @2015 Yes, that would work. Since we've shown that for $Q \in \mathrm{Syl}_{2}(G)$ we have $Q \trianglelefteq G \implies G \text{ abelian}$, by the contrapositive we must have that $G \text{ not abelian} \implies Q \ntrianglelefteq G$. So if $G$ is not abelian, it has no normal Sylow $2$-subgroups. $\endgroup$ – Bill Wallis May 23 '18 at 17:42

Suppose, by way of contradiction, that there is a normal $2$-Sylow subgroup $D$.

If $D$ is cyclic, then it has exactly one involution, and this would form then a conjugacy class made of one element, which does not show up in the class equation. (The "$1$" in the class equation is already taken by the identity.)

This argument actually shows that there is no normal subgroup of order $2$.

If $D$ is a a Klein four-group, it has exactly three involutions, which would then yield a conjugacy class of order $1$ or $3$, again not there.


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