Finding cosets in a quotient group and verify that they are isomorphic to each other

Question: In the group $\mathbb Z_{24}$, Let $H=\langle 4\rangle$ and $N=\langle 6\rangle$.

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

b) List the cosets in $\dfrac{HN}{N}$ showing elements in each coset.

c) List the cosets in $\dfrac H {H\cap N}$, showinng the elements in each coset.

d) Verify that $\dfrac{HN} N$ and $\dfrac H {H\cap N}$ are isomorphic. Definition of isomorphic: Let $G_1$ and $G_2$ be two groups. $G_1$ and $G_2$ are said to be isomorphic if there exist a $\phi : G_1 \to G_2$ such that

a) $\phi$ is a bijection

b) $\phi$ is a homomorphism

I am having problems with this question.
Here is my approach on part a:
$HN=\{0,2,4,6,\ldots,22\}=\langle 2\rangle$

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

Here is my approach on part b:

$\dfrac{HN} N =\dfrac 2 6$

I do not know if part b or a is correct, so I stopped at this point. If anybody can help me, I will appreciate it

• Using $\{\ \}$ simultaneously as the standard notation for subset and as an exotic notation for "smallest subgroup containing stuff" in an exercise which can be summarised in "go from one notation to the other to see if you understand it" literally scraches the bottom of bad ideas. – user228113 Mar 20 '17 at 1:16
• What do you suggest I do? I do not see the correct way to do it @G.Sassatelli – behold Mar 20 '17 at 1:20
• @behold Angle brackets are more common and are not ambiguous. – Qudit Mar 20 '17 at 1:49
• Like the one's I did at the top? – behold Mar 20 '17 at 1:50
• I figured it the angle brackets out. thanks – behold Mar 20 '17 at 2:26

You're fine until you start to deal with $HN/N \stackrel{\text{def}}{=} (H+N)/N$. Let's enumerate the cosets:

$N = \{0,6,12,18\}$

$2+N = \{2,8,14,20\}$

$4+N = \{4,10,16,22\}$.

As you can see, we have $3$ cosets, so $(H+N)/N$ is cyclic of order $3$.

The cosets of $H/(H \cap N)$ are:

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

$4 + (H \cap N) = \{4,16\}$

$8 + (H \cap N) = \{8,20\}$.

This again forms a cyclic group of order $3$. I leave it to you to define the isomorphism.

• I appreciate the help. I can see it now. Thank you – behold Mar 21 '17 at 23:15