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I am trying to understand how $2 \mathbb{Z}$/$6 \mathbb{Z} $ is isomorphic to $\mathbb{Z_3}$ .

So far I understand that:

$2 \mathbb{Z}$/$6 \mathbb{Z} $ = { $0+6\mathbb{Z}, 2+6\mathbb{Z}, 4+6\mathbb{Z}$ }

and that elements of integer mod $3$ are:

$\mathbb{Z}_3$ ={ $0, 1, 2$ }

However, I cannot figure out how $2 \mathbb{Z}$/$6 \mathbb{Z} $ is isomorphic to $\mathbb{Z_3}$. Any help would be very much appreciated.

Sai82
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    One way (in this case, perhaps the way) is to come up with some clever function between the two and show that it is a bijective homomorphism. – Arthur May 24 '14 at 10:30
  • There are lots of good answers here. If you are still struggling then I suggest you write out the two Cayley tables and you will see the connection. – fretty May 25 '14 at 10:54

4 Answers4

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One way is to exhibit an isomorphism directly. So consider the map $\phi:2\Bbb Z/6\Bbb Z \to \Bbb Z_3$ given by:

$\phi(2k + 6\Bbb Z) = k \text{ (mod }3)$

One caveat: when we define a map on an equivalence class by using "representatives" we need to be sure we get the same result no matter which representative we choose. Put another way, we need to show that if for two integers $k,k'$ we have:

$2k + 6\Bbb Z = 2k' + 6\Bbb Z$

then we need to be sure that $\phi$ gives us the same image, that is that:

$k \equiv k' \text{ (mod }3)$.

It should be clear that $2k + 6\Bbb Z = 2k' + 6\Bbb Z$ if and only if $2k - 2k'$ is a multiple of $6$. Now if: $2k - 2k' = 6t$, for some integer $t$, we have that $2(k - k') = 6t$, so that $k - k' = 3t$, which is precisely what it means for $k$ and $k'$ to be congruent modulo $3$.

All this is to be sure that $\phi$ is a "well-defined" function on the cosets, even though we are using elements of the cosets to define the value of $\phi$ at a coset.

With that bit out of the way, the rest is downhill:

$\phi((2k + 6\Bbb Z) + (2m + 6\Bbb Z)) = \phi((2k+2m) + 6\Bbb Z)$

$= \phi(2(k+m) + 6\Bbb Z) = (k+m) \text{ (mod }3) = k \text{ (mod }3) + m \text{ (mod }3)$

$= \phi(2k + 6\Bbb Z) + \phi(2m + 6\Bbb Z)$, so $\phi$ is a homomorphism, and the correspondence:

$0 + 6\Bbb Z \mapsto 0 \text{ (mod }3)$

$2 + 6\Bbb Z \mapsto 1 \text{ (mod }3)$

$4 + 6\Bbb Z \mapsto 2 \text{ (mod }3)$

makes it clear $\phi$ is bijective.

David Wheeler
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Consider the canonical projection $\Bbb Z \to \Bbb Z/3\Bbb Z$. This surjective homomorphism restricts to a homomorphism $2\Bbb Z \to \Bbb Z/3\Bbb Z$ that is still surjective (prove this!). The kernel is $2\Bbb Z \cap 3\Bbb Z = 6\Bbb Z$, since $2$, $3$ are relatively prime. Now apply the first isomorphism theorem.

Ayman Hourieh
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Use this: For any prime p, there is only one possible group structure containing p elements. As both the 2Z/6Z and Z(3) contain 3 elements and 3 is a prime, both are isomorphic to each other. This is essentially because any group of order p (prime) is cyclic. So an isomorphism mapping generator of one group to the generator of another exists.

Nipun Malhotra
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$2\mathbb{Z}$ is cyclic $\Rightarrow 2\mathbb{Z}/6\mathbb{Z}$ is cyclic as quotient of a cyclic group; moreover $$2\mathbb{Z}/6\mathbb{Z} = <2 + 6\mathbb{Z}>$$

Observe that $$3 (2 + 6\mathbb{Z}) = 6 + 6\mathbb{Z} = 0 \ \ \ \text{in} \ \ \ \ 2\mathbb{Z}/6\mathbb{Z}$$

Thus $2\mathbb{Z}/6\mathbb{Z} \cong \mathbb{Z}_3$

WLOG
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