Finding left cosets

Question

Let $G = U(21)$ and $H = \langle 4\rangle \le G$. Find all of the left cosets of $H$ in $G$.

I am able to do the math once I have it set up. Setting it up is difficult and the definition of $\langle 4\rangle$ is a little unclear to me.

Answer 1

<4> is, in this context, all the power of 4. 4^0 = 1, 4^1 = 4 4^2 = 16 = -5 (since we're in Z/21Z). 4^3 = 4*4^2 = 4*(-5) = -20 = 1. Since we got back to unity, we can stop.

Now we just take all the cosets (since G is abelian, we can ignore the "left" part). For each element of G, we multiply it be all the element of <4>:

1 -> {1*1,1*4,1*16} = {1,4,16}
2 -> (2*1,2*4,2*16} = {2,8,11}
etc.

Then you just have to list all the unique sets. That is, list all the sets that result, but don't list the same set twice. So for instance, 1*<4> = {1,4,16} and 4*<4> = {4*1,4*4*,4*16} = {4,16,1}. Since those two sets have the same elements in different order, they are the same set, and you should list this only once.

The higher level way, however, is to think of the structure of G. Since 21 factors into 7*3, Z₂₁ is isomorphic to Z₇ x Z₃. Taking multiplicative units gets rid of the additive identity in both, leaving Z₆ x Z₂ = Z₃ x Z₂ x Z₂. So G is isomorphic to Z₃ x Z₂ x Z₂. The order of 4 is 3, and the orders of 20 and 13 are 2, so G is isomorphic to <4> x <20> x <13>. Thus, to find a cosets, you can just take an element of <20> x <13>, and multiply it by <4>. Taking all four elements of <20> x <13> gives you the four cosets:

(1 x 1) * <4> = {1,4,16}
(1 x 13) * <4> = {13,10,19}
(20 x 1) * <4> = {20,17,5}
(20 x 13) * <4> = {8,11,2}

Answer 2

If $G$ is a finite group and $H$ is a subgroup, then you can find all (left) cosets with the following procedure:

• $H$ is one of the cosets, precisely $g_1H$, where $g_1=1$;
• choose an element $g_2$ in $G$ but not in $g_1H$ and compute $g_2H$;
• choose an element $g_3$ in $G$ but not in $g_1H\cup g_2H$ and compute $g_3H$;
• …
• choose an element $g_k$ in $G$ but not in $g_1H\cup g_2H\cup\dots\cup g_{k-1}H$ and compute $g_kH$.

You know that the process ends at $k=|G|/|H|=[G:H]$ (the index of $H$ in $G$). You always choose an element outside of the previous found ones because you know that distinct cosets are disjoint.

In this case, $H=\langle 4\rangle$ denotes the subgroup consisting of all powers of $4$: $$ H=\{1=4^0,4=4^1,16=4^2\} $$ and we stop here because $4^3\equiv 1\pmod{21}$. Thus, with the procedure outlined above we'll end up with $12/3=4$ cosets, namely

• $g_1=1$, $g_1H=H=\{1,4,16\}$;
• $g_2=2$, $g_2H=2H=\{2,8,11\}$;
• $g_3=5$, $g_3H=5H=\{5,20,17\}$;
• $g_4=10$, $g_4H=10H=\{10,19,13\}$.

The choices of $g_2$, $g_3$ and $g_4$ can be different, but the final list of cosets will be the same, maybe in another order.