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My attempt:

If we take $n$ to be finite, since $\mathbb Z/n\mathbb Z$ is cyclic, there is a distinct subgroup of order $d$ for each divisor $d$ of $n$.

Any subgroup of a cyclic group is cyclic. Any subgroup of order $d$ is isomorphic to $\mathbb Z/d\mathbb Z$, as it is a cyclic group of order $d$.

Cyclic groups are abelian, and $gh=hg$ leads to $gH = Hg$, so all subgroups are normal.

Thus we can view $\mathbb Z/d\mathbb Z$ as the kernel of some homomorphism from $\mathbb Z/n\mathbb Z$ to a group $H$. By the first isomorphism theorem, we have $(\mathbb Z/n\mathbb Z)/(\mathbb Z/d\mathbb Z)$ is isomorphic to the image of that homomorphism.

It is at this point I've become stuck, as I'm not sure what I can say about the image of this homomorphism. I feel like it should be $\mathbb Z/k\mathbb Z$, where $k$ is the integer such that $kd = n$, and this would prove the quotient group is cyclic, as it is isomorphic to a cyclic group. However, I am not sure if this is right or how to justify this?

Any direction is greatly appreciated!

Shaun
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srh
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2 Answers2

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A quotient group $H$ of a group $G$ has a surjective homomorphism $f:G\to H$. If $G$ is cyclic and $x$ is a generator of $G$, then every element of $G$ is $x^n$ for some $n$. Thus the elements of $H$ consist of $f(x^n) =f(x)^n$ since $f$ is surjective. Thus $f(x) $ generates $H$, so $H$ is cyclic.

Matt Samuel
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Any homomorphic image of a cyclic group is cyclic (easy to prove).