# Number of non-identity elements of order $2$ in this set. (Related to abelian group of order $34.$)

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?

• perhaps there is only one element/subgroup of order $2$ – J. W. Tanner Nov 13 '20 at 16:29
• Suppose that $G$ were cyclic of order $34$. What is the answer in that case? – lulu Nov 13 '20 at 16:29
• $G$ is Abelian, has an element of order $2$ and an element of order $17$, hence it has an element of order $34=\mathrm{lcm}(2,17)$. – Mor A. Nov 13 '20 at 16:36
• Cf. this – J. W. Tanner Nov 13 '20 at 16:39

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.