# List the elements and cosets

In the group $\mathbb{Z}_{24}$, let $H=\langle 4 \rangle$ and $N=\langle6\rangle$

a. list the elements in $HN$ (usually write $H+N$ for these additive groups) and $H\cap N$

So I think

$H=\langle4 \rangle= \{0,4,8,12,16,20 \}$ and $N=\langle6 \rangle= \{0,6,12,18 \}$

$H+N= \{0,2,4,6,10,12,14,16,18,20,22 \}.$ I added every element of H to every element of N and obtained these values.

$H\cap N=\{0,12\}$ Since these are the only two commons element in $H$ and $N$.

b. List the cosets in $HN/N$, showing the elements in each coset.

$HN/N=(H+N)/N$

So I am still really confuse about how to find cosets but here's my attempt:

So I need to look at H+N: $H+N=\{0,2,4,6,10,12,14,16,18,20,22\}$

and $N= \langle 6 \rangle$, so I need to separate it by multiples of $6$: $N+0=N+6=N+12=N+18=\{6,12,18\}$

and then $N+2=N+8=N+14=N+20= \{6,14,20,26 \equiv 2\}=\{2,6,14,20\}$

and finally $N+4=N+10=N+16=N+22=\{10,16,22,28 \equiv 4 \}= \{4,10,16,22 \}$

and together these 3 cosets make up $H+N$.

c. List the cosets in $H/(H\cap N)$, showing the elements in each coset.

So I think this is similar to above: $H=\langle 4\rangle =\{0,4,8,12,16,20\}$

$H\cap N=\{0,12\}$

so $H\cap N+0=H\cap N+12=\{0,12\}$

$H\cap N+4=H\cap N+16=\{16,28 \equiv 4\}=\{4,16\}$

$H\cap N+8=H\cap N+20=\{20,32\equiv 8\}=\{8,20\}$

Does this look correct? In particular, am I thinking of cosets correctly?

• You're missing $8$ in $H + N$. Essentially, $H + N$ is the set of even residues modulo $24$; it's isomorphic to $\Bbb Z_{12}$.
Looking at $HN/N$:
• You're missing $0$ in $N$, when you calculate the first coset of $N = 0 + N$.
• Somehow you have $6$ in $N + 2$, when you want $2 + 6 = 8$ (everything in $N + 2$ should be congruent to each other modulo $6$; they should differ by a multiple of $6$).
Everything looks good in $H/(H \cap N)$. In particular, you seem to be thinking of cosets correctly; they're all just subsets of $\Bbb Z_{24}$ here, more specifically 'translates' of certain subgroups (perhaps with your focus restricted to certain subgroups of $\Bbb Z_{24}$, but the idea is the same).