Number of non-identity, self-inverse elements in an Abelian group of order 10 Let G be an Abelian group of order 10. Let $S=\{g \in G : g^{-1}=g\} $. then number of non identity elements in S is
A.$5$
B.$2$
C.$1$
D.$0$
I take group to be addition modulo 10. I get answer C. However i am not sure this is right way
Thanks
 A: Note: $S$ consists of elements of order 2. If $G$ is an abelian group of order pq, distinct primes then $G \cong \mathbb{Z}_{p} \times \mathbb{Z}_{q}$
A: The non identity elements of $S$ are the elements of $G$ of order $2$. Therefore, we distinguish the following two cases:


*

*Case 1: If G is cyclic then the number elements of order $2$ are $\phi(2)=1$ where $\phi$ is the Euler function.

*Case 2: If G is not cyclic then the answer is again $1$.

*Proof of Case 2: If there were two elements $a,b\ (a\neq b)$ of order $2$ in $G$ then due to the fact that $G$ is abelian we would have that the set $$H:=\{a,b,ab,e_G\}$$ would form a subgroup of $G$ which is a contradiction, due to Lagranges theorem.  ($4=\text{o}(H)\nmid \text{order}(G)=10)$


PS: Every group of even order has a non trivial element of order two! 
A: $S$ is a sub-group of $G$ because $G$ i Abelian, so $|S|$, the number of members of $S,$ is a divisor of $10.$  So $|S|\in \{1,2,5,10\}.$ Minus the identity, we have $|S|-1\in\{0,1.4,9\}.$ You are told that $|S|-1\in \{0,1,2,5\}.$ So $|S|-1\in \{0,1\}.$ We can eliminate the case $|S|-1=0$ as follows : If $|S|-1=0$ then every $x\in S$, which is not equal to the identity $e$,  has order $2,5$,or $10.\quad$(  $[x],$ the order of $x,$ is the least $n\in N$ such that $x^n=e.$ The set $\{x^j: 0\leq j\leq [x]-1\}$ is a sub-group, so $[x]$ is a divisor of $10.$) But if $[x]=2$  for any $x$ then $|S|-1>0.$ And if $[x]=10$  for any $x$ then $[x^5]=2,$ giving $|S|-1>0.$ And we cannot have every $x\in S$ except $e$ having order $5,$ for if we let $H_x=\{x^j :0\leq  j\leq 4\}$ for any $x\ne e$, then for any $y\in G\backslash H_x $ we have  $G=H_x\cup y H_x.$ But now $y^2\in y H_x\implies y=y^{-1} y^2\in y^{-1} y H_x=H_x,$ which is false. So $y^2\in H_x.$ Since [y]=5,this gives $y=e y=e^3 y=(y^5)^3  y=y^{16}=(y^2)^8\in H_x,$ a contradiction.
A: Note that $S$ is a subgroup of $G$ and consists of the elements $g$ such that $g^2=1$; by Lagrange's theorem we can only have $|S|=1$ or $|S|=2$; indeed, by Cauchy's theorem, a group of order $10$ has elements of order $5$, so $S\ne G$, which dismisses the case $|S|=10$; the same argument dismisses the case $|S|=5$.
Since, by the same theorem, $G$ has at least an element of order $2$, the case $|S|=1$ cannot hold.
