# Normal Complement to a Sylow $2$-subgroup in a group of order $4$ mod $8$.

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.

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

Hint: take a look at Burnside's normal $$p$$-complement theorem.