A Group Theory question involving isomorphism and cyclic groups

Show that if G is a group of order 4 then either G is isomorphic to the cyclic group Z4 of order 4, or x^(2)=1 for all x in G.

I have got this far....

For two groups to be isomorphic they must have the same order, be cyclic and be abelian.

Case 1: G is cyclic (and abelian) therefore is isomorphic to the cyclic group Z4

Case 2: G is not cyclic (not abelian)............therefore x^(2)=1 for all x in G.

I don't understand the connection between the group not being cyclic or abelian and the condition x^(2)=1.

Please help, thanks.

Quote:

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Originally Posted by Louise

Show that if G is a group of order 4 then either G is isomorphic to the cyclic group Z4 of order 4, or x^(2)=1 for all x in G.

I have got this far....

For two groups to be isomorphic they must have the same order, be cyclic and be abelian.

Case 1: G is cyclic (and abelian) therefore is isomorphic to the cyclic group Z4

Case 2: G is not cyclic (not abelian)............therefore x^(2)=1 for all x in G.

I don't understand the connection between the group not being cyclic or abelian and the condition x^(2)=1.

Please help, thanks.

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If G is cyclic => G is isomorphic Z4
If G is not cyclic => no element has order = 4. => every element has order 1 or 2. (as these are only two numbers that divide 4)
hence x^2 = 1 for all x in G.