This question was asked in a bachelor exam for which I am preparing and I was unable to solve it.
Let $G$ be an abelian group of order $34$ and $S=\{ g \in G \mid g=g^{-1}\}$ . Then what is the number of non- identity elements in $S$?
I used sylow theorem: There is $1$ sylow subgroup of order $17$ and $17$ sylow subgroups of order $2$ but $17$ is not the answer ( I'm not even close!) .
What is wrong in my approach? Can you please tell?
In a finite abelian group $G$, there's only one $p$ Sylow subgroup for all $p$. This follows from the Sylow theorems: all $p$ Sylow subgroups are conjugate, but in an abelian group, conjugation doesn't do anything. Thus an abelian group of order $34=2 \cdot 17$ will have one $2$-Sylow subgroup and hence one non-identity element $g \in G$ satisfying $g=g^{-1}$.
Hint: Since $\lvert G\rvert=34=17\times 2$, by the classification theorem for groups of order $2p$ for prime $p$ greater than two,${}^\dagger$ we have that either $G\cong \Bbb Z_{2p}$ or $G\cong D_p$, the dihedral group of order $2p$. (Let $p=17$.)
But $G$ is abelian, so . . .
$\dagger$: Theorem 7.3 of Gallian's "Contemporary Abstract Algebra (Eighth Edition)".
By the classification theorem for finite abelian groups, $G$ is cyclic of order 34. A finite cyclic group of order $n$ has exactly one subgroup of order $d$ for any $d\mid n$, so there is exactly one subgroup of order two. Since any element of order two generates a subgroup of order two, there is precisely one such element. So there are two elements with $g=g^{-1}$: the identity and the unique element of order two.
