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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?

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    $\begingroup$ Related, if not a duplicate. $\endgroup$ May 6 at 3:39
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    $\begingroup$ 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. $\endgroup$ May 6 at 3:48
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    $\begingroup$ I would advise you to consider the number of Sylow $5$-subgroups. $\endgroup$
    – Derek Holt
    May 6 at 7:47

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