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For a group of order $n$ congruent to $4$ mod $8$, the Sylow $2$-subgroup has order $4$, and hence is either cyclic, or elementary abelian.

In the first case (cyclic Sylow) we know that there is a normal complement, and clearly in the other case there may not be. An example of course is the alternating group $A_5$, or any of the other simple groups of order $4$ mod $8$. But all of the simple groups of that size $3$ divides the order of the group.

Is it possible that if $3$ does not divide the order of the group, then we do get a normal complement? Does anyone have a counter example, or a reference. Edit: Thanks for the hints. Burnsides theorem works fine if the sylow 2 group is cyclic, but it needs help if the sylow 2 subgroup is the non-cyclic group of order 4. That is the reason for the hypothesis that 3 does not divide the order of the group. As of now I do not see how to apply this. As pointed out below by DH and JL, the key is that the action of the normalizer N of the sylow subgroup on itself gives a homomorphism to the automorphism group of C_2 x C_2, which is S3. The kernel of this homomorphism is the center of N, so N/Center(N) injects into S3. But N has no elements of order 3, since G doesn’t.

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    $\begingroup$ The answer to your question is yes, and as Andreas Caranti points out, you can prove it using Burnside's Transfer Theorem. $\endgroup$ – Derek Holt Aug 20 '19 at 10:30
  • $\begingroup$ To spell it out: the automorphism group of $C_2\times C_2$ is isomorphic to $S_3$. $\endgroup$ – Jyrki Lahtonen Aug 21 '19 at 3:45
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Hint: take a look at Burnside's normal $p$-complement theorem.

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