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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

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    $\begingroup$ For sets use \{ not {. $\endgroup$
    – David
    Jan 20, 2016 at 5:28
  • $\begingroup$ @David ok thanks $\endgroup$
    – Taylor Ted
    Jan 20, 2016 at 5:32
  • $\begingroup$ Granted that you already know that it has to be one of the four possible answers your argument works. $\endgroup$ Jan 20, 2016 at 5:35
  • $\begingroup$ What about having the four possible answers allows the assumption that $G = \mathbb{Z}/10\mathbb{Z}$? $\endgroup$ Jan 20, 2016 at 5:42
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    $\begingroup$ @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. $\endgroup$ Jan 20, 2016 at 5:50

4 Answers 4

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

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  • $\begingroup$ I'm not a mathematician and I fail to see how this answers the question. $\endgroup$
    – rubenvb
    Jan 21, 2016 at 17:42
  • $\begingroup$ @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. $\endgroup$
    – Elliot
    Jan 22, 2016 at 3:29
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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!

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

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  • $\begingroup$ 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). $\endgroup$ Jan 20, 2016 at 6:43
  • $\begingroup$ Right. 4 does not divide 10.Thanks. $\endgroup$ Jan 20, 2016 at 15:25
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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.

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