The centralizer of a subgroup is the set of elements that commute with everything in the subgroup. Verify that the centralizer of H is a group. Apply the a/b test. Inverse is the only tricky part. If b commutes with H, does b inverse? Given H/b, multiply by 1 = (1/b)*b on the left. This simplifies to 1/b times H.
The center is the centralizer of the entire group. In other words, the center is the subgroup that commutes with all of G. The center is always a normal subgroup.
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The normalizer of a subgroup H is the set of all elements x such that xH/x = H.
Clearly H is in the normalizer of H.
Remember that conjugation is an inner automorphism, 1-1 and onto.
When x comes from the normalizer of H,
xH/x implements an automorphism on H.
This can be reversed, hence 1/x implements an automorphism on H,
and 1/x is in the normalizer.
Similarly, xy is in the normalizer when x and y are both in the normalizer.
Thus the normalizer is a group.
Since everything in the normalizer fixes H by conjugation, H is normal in its normalizer. Don't confuse the normalizer of H with the normal subgroup generated by H. They are quite different. When H is normal in G it is its own normal subgroup, yet its normalizer is all of G. Conversely, the smallest normal subgroup containing H might be all of G, and the normalizer of H could simply be H. |
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