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In any group G, some of the elements of G commute with all of the other elements in G
In any group G, some of the elements of G commute with all of the other elements in G.
The set of all such elements in G is called the centre of G, and is denoted by Z(G).
Hence Z(G) = {g|xg = gx x is a subset of G}.
For instance in any group the identity commutes with every element - so Z(G) is never empty. It should also be clear that:
Z(G) = G if and only if G is Abelian
Prove that this is always true. i.e Prove that in any group G, the centre Z (G) is a subgroup of G. In fact, Z (G) is always a normal subgroup of G (but you are not asked to prove this)
Expert Solution
Problem #4
(a) I show that Z(G)=G if and only if G is abelian.
"=>": If Z(G)=G, then for any a,b in G, a and b are also in Z(G).
By the definition of Z(G), we have ab=ba and thus G is abelian.
"<=": If G is abelian, we select any element x in G. Then for any
element y in G, we have xy=yx. By the definition of Z(G), x should
be in Z(G). Thus G is contained in Z(G). But Z(G) is also contained
in G, thus we must have Z(G)=G.
(b) I show that Z(G) is a subgroup of G.
For any a,b in Z(G), x in G, we have ax=xa, bx=xb.
From bx=xb, we have bxb^(-1)=x, then xb^(-1)=b^(-1)x
Thus we have
ab^(-1)x=a(b^(-1)x)=a(xb^(-1))=(ax)b^(-1)=xab^(-1)
So ab^(-1) is also in Z(G).
Thus Z(G) is a subgroup of G.
More over, Z(G) is normal. Because for any a in Z(G), x in G, we have
xax^(-1)=axx^(-1)=a in Z(G), thus xZ(G)x^(-1)=Z(G) and hence Z(G) is normal.
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