1. Isomorphism Question

G has 16 elements and H has 4 elements....so G/H = 4?

Now I have what seems like a corollary in my book, but not stated as...it says "a group of order 4 is isomorphic to Z4 or Z2+Z2 (direct prodcut)"...so would the first part be yes? Maybe I am over complicating things...
and then on the second part
k = <(1,2)>
How do you find out how many elements a generating function like K has?

2. Originally Posted by xyz

G has 16 elements and H has 4 elements....so G/H = 4?

Now I have what seems like a corollary in my book, but not stated as...it says "a group of order 4 is isomorphic to Z4 or Z2+Z2 (direct prodcut)"...so would the first part be yes? Maybe I am over complicating things...
and then on the second part
k = <(1,2)>
How do you find out how many elements a generating function like K has?
I'm sorry. But do you know what $G/H$ means?

3. Let G be a group and let H be a normal subgroup of G.
The set G/H = {aH | a ∈ G} is a group under the operation (aH)(bH) = abH. I mentioned that the order of G/H is 4.

4. what makes $\mathbb{Z}_2 \oplus \mathbb{Z}_2$ and $\mathbb{Z}_4$ different is that $\forall x \in \mathbb{Z}_2 \oplus \mathbb{Z}_2: \ 2x=0,$ which is not true in $\mathbb{Z}_4.$ now we have $K=\{(0,0),(1,2),(2,0),(3,2) \}.$ so if you let $x=(3,1) + K \in G/K,$ then

$2x=(2,2) + K \neq 0,$ because $(2,2) \notin K.$ thus $G/K \cong \mathbb{Z}_4.$ such situation does not happen in $G/H,$ which means $G/H \cong \mathbb{Z}_2 \oplus \mathbb{Z}_2.$