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
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.