The normalizer in G of a subgroup H of G consists of all of

The normalizer in G of a subgroup H of G consists of all of the elements of G which fix the subgroup H under conjugation. BEWARE: this is not the same thing as fixing all of the elements of H under conjugation, that would refer to the-centralizer. An element a is in the normalizer provided that a fixes the set H under conjugation. So a does not move the set H to a different subgroup. For Your Understanding: Given G = Q_s and H = , compute N_G(H).

Solution

G=Q₈ is the quaternion group

H =<i>, is the subgroup generated by i.

So H={1,-1,i ,-i}

Method 1

As H is a subgroup of index 2 in G, it is normal in G

So N_G(H) = G .(ie , every element of G normalizes H).

Method 2:

We can directly prove that G normalizes H by considering

jij^-1 =-k(-j) = -i which is in H.

Thus j and hence -j normalize H. SImilar proof for k.

Hence , again , N_G(H) = G .