# Exhibiting the distinct cosets of a cyclic subgroup H in group G?

Let G be the group U(24) and H the cyclic subgroup generated by the element 7.

Exhibit all of the distinct cosets of H in G.

I am coming up with this. (I created the Cayley table for U(24) - using the elements 1,5,7,11,13,17,19,23.)

<1> = {1} <5> = (5*5 = 1) {5} <7> = (7*7 = 1) {7}

and so forth since each element times itself is equal to the identity.

Is this correct? there are 8 elements and therefore 8 distinct cosets for U(24).

Next piece of the problem I'm working thru.

Why is H a normal subgroup of G?

I am not quite sure what H is... exactly to compare

I'm guessing <7> = H

so by looking at the left and right cosets

<1>H <-> H<1> <5>H (5*7 = 11) <-> H<5> (7*5 = 11) <7>H (7*7 = 1) <-> H<7> (7*7 = 1)

and then I would stop there and say it was normal subgroup of G. with order 1 and the index of H in G is 2?

But Lagrange Theorem states the order of G = (order of H)* index of H in G... this doesn't end up working in this case (Makes me wonder if I am doing this part wrong)

Exhibit distinct elements of the quotient group of G/H... (So I can construct a Cayley Table for G/H.)

I am just starting to work on this... some help to get me started would be nice :)

Thanks

-

The group $H$ has two elements $\{7, 1\}$ and therefore its index in $G$ is 4. Also note that right cosets coincide with left cosects so that $gH = Hg$ (this comes from the fact that G is abelian and also answers your second question). So e.g. $5 \{7, 1\} = \{11, 5\} = 11 \{7, 1\}$. You'll get 4 cosets like this.
Third question is really the same thing as the first question (thanks to $H$ being normal).