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Let $ G= \mathbb{Z^3}$ and consider $ N=\{(i,j,k) \in \mathbb {Z^3} : i+2j=3i-k=0 \} $ Assuming N is a normal subgroup of G find a familiar group H s.t $ G/N \cong H$

I started by writing i want to find a $\phi $ where $ \ker \phi(i,j,k) = (0,0)$ iff $i+2j =0$ and $ 3i-k=0 $ i belive i want $ \phi(i,j,k) = ( (i+2j), (3i-k)) $ where $\phi: G \to \mathbb{Z^2} $

$\phi$ is a homomorphism attempt:

$ \phi(i,j,k) \phi (a,b,c) = ((i+2j) + (a+2b),(3i-k) + (3a-c)) = \phi(i+a,j+b,k+c) $ so phi is a homomorphism. im not sur ei put these together right?

i believe by construction $\ker \phi = (i+2j, 3i-k)$ so i need to somehow show that this is the same as N?

Faust
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1 Answers1

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Your construction is nice, now you to prove that $N = \ker \phi$ isn't hard, as for every $g \in N$ we have that $\phi(g) = (0,0)$, so $N \subseteq \ker \phi$. The other relation is also trivial and so $N = \ker \phi$

Now we have that: $G/N = G/\ker \phi \cong \phi[G]$. Now it's not hard ot prove that the image of $G$ is $\mathbb{Z} \times \mathbb{Z}$, as $\phi$ maps $G$ to $\mathbb{Z} \times \mathbb{Z}$ and for each element of $\mathbb{Z} \times \mathbb{Z}$ we can find a pre-image in $G$.

Stefan4024
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  • so say i want to show that $\ker \phi \subset N $ i cant seem to figure out what element i am picking cause they look the same to me? for the second part i think you mean $G/N = G/ \ker \phi \cong \phi (G) $ ? then for eveyr pair of numbers (a,b) there exists and i,j,k s.t i+2j= a and 3i-k = b where a,b,i,j,k are integers??? which may imply that $im(G)$ is onto $\mathbb{Z}$ x$ \mathbb{Z} $ ? – Faust Oct 15 '17 at 20:43
  • @Faust As I said the relation is trivial. For example take an element of $N$. It's $(i,j,k)$ s.t. $i+2j=3i-k=0$. Then we have that $\phi(i,j,k) = (i+2j,3i-k) = (0,0)$. For the other direction let $(i,j,k) \in \ker \phi$ then we have that $i+2j=3i-k=0$, which exactly describes the subgroup $N$. – Stefan4024 Oct 15 '17 at 20:48
  • @Faust Yeah I indeed meant using $\cong$ sign. For the last question you're the right track. Prove that $\phi$ is onto $\mathbb{Z} \times \mathbb{Z}$, which can be done by solving $i+2j = a$ and $3i-k=b$ in terms of $a,b$. – Stefan4024 Oct 15 '17 at 20:50
  • I think i follow ok, but i still dont understand $G/H $ i thought $\phi : G \to H $ ? – Faust Oct 15 '17 at 20:50
  • Oh i think is just a typo i think you meant $G/N $ if so i think im understanding thanks =) – Faust Oct 15 '17 at 20:52
  • @Faust Oh yeah, you're right. – Stefan4024 Oct 15 '17 at 21:03