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Do you have an idea how to tackle this problem:

The center of a finite group G whose order is 44, has an element of order 2.

I believe the idea must somehow involve the Sylow's Theorem; for example the Sylow 11-subgroup, say S, is normal. If we put C:=C(S) then C:=C(S)≤N(S)=G. We have that G/C is isomorphic to a subgroup of Aut(S), which has order φ(11)=10. Thus the order of G/C divides both 10 and 44; hence it could be either 1 or 2. No more progress yet!

In fact I had already asked a quite similar question, but the same method does not seem to work,,,

Thank you for considering this problem!

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You're very close. If the order of $G/C$ is $1$, then the Sylow $11$-subgroup of $G$ is central. If I consider any element of order $2$, then its centralizer will contain a subgroup of order $4$ and the Sylow $11$-subgroup. (A group of order $4$ is always abelian so all Sylow $2$-subgroups of $G$ are abelian.) However, this implies the element in question is central. (Incidentally, this proves that $G$ is abelian if the Sylow $11$-subgroup is central.)

Can you apply a similar argument to the case where the order of $G/C$ is $2$?

I hope this helps!

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