# 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

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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).

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So the Cayley Table for the third part is really the same as the U(24) table OR am I using the 4 cosets as the elements in the table? I have a feeling it is taking 1H, 5H, 13H and 17H and building a Cayley table with those distinct elements. –  spittballz Nov 8 '10 at 8:46
I see now that order G = 8, order H = 2 and the index of H in G is now 4 --- so that does work itself out. –  spittballz Nov 8 '10 at 8:46
Back to first comment... trying to work that part out. 1H being the identity, all distinct elements times itself gives 1H, but then for example what would 5H*13H become? (11,5)*(19,13) = (19,13,11,5) that result is not 1H,5H,13H or 17H. –  spittballz Nov 8 '10 at 8:54
How would I describe this group which is isomorphic to G/H and thus being a nontrivial homomorphic image of U(24)? Using in the description one of the following: cyclic order of..., non-cyclic abelian of order..., or nor-abelian of order...? Not sure what this is asking and how I can answer it. –  spittballz Nov 8 '10 at 8:58
Ad third comment: recall that group operation on G/H is defined as aH * bH = (a*b)H (this is well-defined precisely because H is normal). E.g. 5H * 13 H = (5*13)H = 17H. –  Marek Nov 9 '10 at 10:48
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