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Homework Statement
Prove that the center of the factor group G/Z(G) is the trivial subgroup ({e}).
Homework Equations
Z(G) = {elements a in G|ax=xa for all elements x in G}
The Attempt at a Solution
I need to prove G is abelian, because G/Z(G) is cyclic, right?
Then I can say that since G is abelian, Z(G) = G and G/Z(G) is trivial. So the center of the trivial factor group is the trivial subgroup ({e})?
This is my argument, but I am not sure if it is even headed in the right direction.
If G/Z(G) is cyclic, then by an element g in G, the generator gZ(G) allows gZ(g)= G/Z(G). Suppose elements a,b in G exist and aZ(G) = (gZ(G))^i = g^i * Z(G)
and bZ(G) = (gZ(G))^j = g&j * Z(G).
If a=g^i * x
and b= g^j * y
then, ab = (g^i * x) (g^j * y) = g^(i+j) *(xy) = ((g^j)y)((g^i)x) = ba
Since ab=ba , G is abelian.
By definition of an abelian group G, and given that Z(G)= {elements a in G|ax=xa for all elements x in G}, we conclude that Z(G) = G. Thereby, if ax=xa, for all elements x in G, then x must be the identity element e. The factor group G/Z(G) contains a single element: the identity element e, which is known as the trivial subgroup. (This last part isn't formalized, how should I go about doing that?)