# Prove that groups of order 200 are solvable

Suppose $$G$$ is a group with $$|G| = 200$$. Since $$200 = 2^3 \cdot 5^2$$ I've used Sylow's theorem to make two claims concerning $$n_G(8)$$, the number of Sylow 2-subgroups of $$G$$:

$$n_G(8) \equiv 1 \mod 2, \quad n_G(8) \mid 25$$

The first claim gives that $$n_G(8) \in \{1, 3, 5, 7, ...\}$$ and the second gives that $$n_G(8) \in \{1, 5, 25\}$$ so together $$n_G(8) \in \{1, 5, 25\}$$.

I think I've solved the case for when $$n_G(8) = 1$$, in which the unique Sylow 2-subgroup $$P$$ is normal in $$G$$, so I can construct the quotient group $$G/P$$. $$P$$ has order 8, while $$G/P$$ has order $$200/8 = 25$$, so both $$P$$ and $$G/P$$ are $$p$$-groups and hence solvable, hence $$G$$ is solvable.

I am stuck about what to do in the case that there is either 5 or 25 Sylow 2-subgroups, since in either of these instances the Sylow 2-subgroups will not be normal in $$G$$. How does one go about proceeding in this instance?

• Related, if not a duplicate. May 6 at 3:39
• Perhaps you can try a counting argument to show that you may not have 5 or 25 Sylow 2-subgroups in $G$, for otherwise the cardinality of $G$ would exceed 200. May 6 at 3:48
• I would advise you to consider the number of Sylow $5$-subgroups. May 6 at 7:47