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Let $S$={$5$,$1/5$,$1$} be a set then I think $(S, .)$ is a group where identity element is $1$.Here order of the group is $3$.What is the order of the elements i.e $5$ and $1/5$? We know that the order of each element in a finite group is a divisor of the group. So the order of $5$ and $1/5$ must be $3$ or $1$.But $5^3$ is not $1$. what is my mistake here? Please someone help me.thanks for your kind help.

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Your set $S$ is not a group under multiplication, since it is not closed under multiplication.

(Note that $5^k \notin S$ for any integers $k \notin \{-1, 0, 1\}$.)

Since $S$ is not a group, it makes little sense to discuss the order of the elements in $S$.

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  • $\begingroup$ If $G$ be a group and $a\in G$.then {$a$,$a^{-1}$,$e$}does not form a subgroup of $G$ ? $\endgroup$
    – liesel
    Apr 21 '14 at 19:56
  • $\begingroup$ That depends on the group operation, the set, and on the element $a$. If the operation is multiplicative on the set of positive reals, then no. If addition on integers modulo 6, then yes if $a = 2$ but no if $a = 5$. $\endgroup$
    – amWhy
    Apr 21 '14 at 20:02

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