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Show that $\langle 3\rangle / \langle 12\rangle$ is isomorphic to $\Bbb Z_4$

Can someone help me prove this? I have difficulty with isomorphism. I need to generalise it to any integers $n,k$. If I can show this special case, I may solve it for any integers

Thanks in advance!

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    $\begingroup$ To start, a quotient of a cyclic group is cyclic since the class of a generator is still a generator, and cyclic groups of a given order are unique up to isomorphism. $\endgroup$ – Ben West Apr 6 '16 at 18:13
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Let $G = \langle 3 \rangle$. An arbitrary element of $G$ is of the form $3n$ for some integer $n$.

Define the map $\phi : G \to \mathbb Z_4$ by $\phi(3n) = [n]$, where $[n]$ denotes the equivalence class of $n$, modulo $4$.

First we note that $\phi$ is well-defined, because if $3n = 3m$ then $n = m$, so $[n] = [m]$.

Now we verify that $\phi$ is a homomorphism:

$$\phi(3n + 3m) = \phi(3(n+m)) = [n+m] = [n]+[m] = \phi(3n) + \phi(3m)$$

It's clear that $\phi$ is surjective. Now what is the kernel of $\phi$? It is the set of all $3n \in G$ such that $\phi(3n) = [n] = [0]$. Since $[n] = [0]$ holds if and only if $n$ is a multiple of $4$, the kernel of $\phi$ is precisely the set of multiples of $12$, in other words $\langle 12 \rangle$.

The first isomorphism theorem now allows us to conclude that $G / \langle 12 \rangle \cong \mathbb Z_4$.

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  • $\begingroup$ This is a great answer, thanks! $\endgroup$ – Frank Booth Apr 6 '16 at 18:57

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