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Prove that there is no group that has exactly three elements of order 3.

Please can someone kindly put me through how to prove this?

I know that a group of three elements = {x,y,z}. but proving that no group of exactly three elements exists is my problem.

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    $\begingroup$ AN ELEMENT OF ORDER 3 IS DIFFERENT OF HIS INVERSE. $\endgroup$
    – Problemsolving
    Oct 24, 2016 at 15:30
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    $\begingroup$ The question doesn't ask for a group with three elements (there's the cyclic group $C_3$), but for a group of unspecified size where exactly three of the elements have order $3$. $\endgroup$
    – hmakholm left over Monica
    Oct 24, 2016 at 15:32
  • $\begingroup$ Be careful with your definitions here. The order of an element of a group is not the same as the order of the group. The order of an element g is the smallest positive integer k such that g^k = e. That is different from the order of the group, which is the number of elements it contains. $\endgroup$
    – Nasenhaar
    Oct 24, 2016 at 16:03

3 Answers 3

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Hint: An element has exactly the same order as its inverse. And an element of order $3$ is not its own inverse.

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  • $\begingroup$ can i say <x> = {e, x, x^-1} as a group of order three $\endgroup$
    – Atinuke
    Oct 24, 2016 at 16:13
  • $\begingroup$ @Atinuke: It is a group of order three, but only two of its three elements have order three. $\endgroup$
    – hmakholm left over Monica
    Oct 24, 2016 at 16:15
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Try to make the multiplication table for a group of three elements, it should not be difficult. Think about the inverse for each element as well as each of its powers.

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  • $\begingroup$ pls put me through the table pls. $\endgroup$
    – Atinuke
    Oct 24, 2016 at 16:13
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Every group has the identity element $e$ which always has order $1$ so there goes that.

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