Third Isomorphism theorem

Verify the third isomorphism theorem ($\frac{G/N}{H/N} \cong G/H$ ) for $G=\mathbb{Z}\oplus \mathbb{Z}$, $N=\langle(1,0)\rangle$ and $H= \langle(1,0),(0,5)\rangle$. But how does for example the quotient group $G/N$ look like. Is it {{$\langle(1,0)\rangle$},{$\langle(0,1)\rangle$},{$\langle(1,1)\rangle$,...}?

We can write: $N = \Bbb Z \oplus \{0\}$, and $H = \Bbb Z \oplus 5\Bbb Z$.

So $G/N = \dfrac{\Bbb Z \oplus \Bbb Z}{\Bbb Z \oplus \{0\}}$,

and $H/N = \dfrac{\Bbb Z \oplus 5\Bbb Z}{\Bbb Z \oplus \{0\}}$.

Elements of $G/N$ look like this: $(0,y) + N$, because for any given $y \in \Bbb Z$, we have $(0,y) \in (x,y) + N$, since $(x,y) - (0,y) = (x,0) \in N$.

Elements of $H/N$ look like this: $(0,5y) + N$.

Now $G/H = \dfrac{\Bbb Z \oplus \Bbb Z}{\Bbb Z \oplus 5\Bbb Z}$, and:

$G/H = \{H,(0,1) + H, (0,2) + H, (0,3) + H,(0,4) + H\}$ which is a cyclic group of order $5$.

These are the only $5$ cosets we get: if $(x,y) \in G$ writing $y = 5k + r$ for $r \in \{0,1,2,3,4\}$, we have:

$(x,y) - (0,r) = (x,5k + r) - (0,r) = (x,5k) \in H$.

It's cumbersome to write down explicitly what the cosets of $(G/N)/(H/N)$ are, but they look like this:

$[(0,y) + N] + H/N$.

Note we only get $5$ cosets here, as well, since $[(x,y) + N] + H/N = [(0,r) + N] + H/N$ (for $y = 5k + r$ as above) because:

$[(x,y) + N] - [(0,r) + N] = [(x,5k) + N] \in H/N$ (recall $(x,5k) \in H$).

• Thanks a lot for explaining it very clear!!! I think I understand it now, Thanks!!! – bob Oct 17 '15 at 21:51

Just translate the general context: $$G/N=\bigl\{g+N\mid g\in G\bigr\},$$ hence $$\mathbf Z\times\mathbf Z/\langle(1,0)\rangle=\bigl\{(x,y)+n(1,0)\mid n\in\mathbf Z\bigr\}=\bigl\{(x+n,y)\mid n\in\mathbf Z\bigr\} =\bigl\{(x,y)\mid x\in\mathbf Z\bigr\}.$$ So congruence classes bijectively correspond to the (common) second terms of the pairs in the classes.

Similarly, $H/N=\bigl\{(x,5y)+n(1,0)\mid n\in\mathbf Z\bigr\}=\bigl\{(x,5y)\mid x\in\mathbf Z\bigr\}$.

• Ok thanks, and $\frac{G/N}{H/N}$? I think $H/N =$ {$(x,5y) |x,y \in \mathbb {Z}$} – bob Oct 17 '15 at 19:07
• Is it just {$(x,y) |x,y \in \mathbb{Z}$ }? – bob Oct 17 '15 at 19:10
• And $G/H$ is the same, so they are isomorphic? – bob Oct 17 '15 at 19:11
• You're right, $H/N=\bigl\{(x,5y)\mid x\in\mathbf Z\bigr\}$, so $(G/N)/(H/N)$ is isomorphic to $\mathbf Z/5\mathbf Z$. – Bernard Oct 17 '15 at 20:06