Let us prove that a normalizer N(a)={g∈G:ga=ag} of a is a subgroup of G.
Let h,k∈N(a). Then ha=ah and ka=ak. It follows that
(hk)a=h(ka)=h(ak)=(ha)k=(ah)k=a(hk),
and hence hk∈N(a).
It also follows from ha=ah that a−1haa−1=a−1aha−1, and hence a−1h=ha−1. We conclude that a−1∈N(a).
Therefore, N(a) is a subgroup of G.
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