# 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

• For sets use \{ not {. Jan 20, 2016 at 5:28
• @David ok thanks Jan 20, 2016 at 5:32
• Granted that you already know that it has to be one of the four possible answers your argument works. Jan 20, 2016 at 5:35
• What about having the four possible answers allows the assumption that $G = \mathbb{Z}/10\mathbb{Z}$? Jan 20, 2016 at 5:42
• @AdamFrancey: If we know that one of the four answers is right and because no further assumptions are made about $G$ we can conclude the right answer must hold for every group of order $10$. So we can just take $\mathbb{Z}/10$ to find out which one is the right. This is a meta argument about the way the exercise is presented, and no mathematical argument. And it only works if we assume that one of the answers is right. Jan 20, 2016 at 5:50

Note: $S$ consists of elements of order 2. If $G$ is an abelian group of order pq, distinct primes then $G \cong \mathbb{Z}_{p} \times \mathbb{Z}_{q}$

• I'm not a mathematician and I fail to see how this answers the question. Jan 21, 2016 at 17:42
• @rubenvb at the risk of entirely giving away the answer, I chose to leave these true but maybe slightly cryptic hints that would result in the answer. G in this situation falls under these considerations where with this isomorphism, it is a trivial task to count the elements of order 2. Jan 22, 2016 at 3:29

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!

• There aren't two cases: an abelian group of order ten is cyclic. Jan 20, 2016 at 16:18
• I know! i just didnt want to use the result $\mathbb{Z}_p\times\mathbb{Z}_q$ cyclic iff $(p,q)=1$. Jan 20, 2016 at 16:22

$S$ 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.

• No abelian group of order $10$ has three non-identity elements of order $2$ (nor any other group of order $10$ for that matter). You may also want to mention why $S$ is a subgroup (this uses that the group is abelian). Jan 20, 2016 at 6:43
• Right. 4 does not divide 10.Thanks. Jan 20, 2016 at 15:25

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.