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Let $G=\Bbb Z/n \Bbb Z$ and assume $d\mid n$.

There exists a unique subgroup of $G$ of order $d$.

The mod $n$ reduction of $n/d\in \Bbb Z$ generates a $d$ element subgroup of $\Bbb Z/n\Bbb Z$, so that takes care of existence.

What about uniqueness?

If $H\leq G$ is any subgroup of order $d$, then $G/H$ has order $n/d$, right? What result did I use here?

In particular $n/d\cdot g\in G/H$ is trivial, so $n/d\cdot g\in H$ for $g\in G$. Taking $g=1$ we have $n/d\in H$ and there are $d$ multiples of this in $H$, so $H$ is the same group as above. Does this argument work?

MyNameIs
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3 Answers3

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Consider the canonical map $\pi\colon\mathbb{Z}\to G=\mathbb{Z}/n\mathbb{Z}$ and let $d$ be a divisor of $n$; set $n=dk$.

Existence. Consider the subgroup $k\mathbb{Z}$. Then $\pi(k\mathbb{Z})$ is a subgroup of $G$ and it has $d$ elements (prove it).

Uniqueness. Suppose $H$ is a subgroup of $G$ having $d$ elements; then $\pi^{-1}(H)$ is a subgroup of $\mathbb{Z}$ containing $n\mathbb{Z}$; it's easy to see that necessarily $\pi^{-1}(H)=k\mathbb{Z}$.

egreg
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  • Thanks!! Could you expand on the last sentence? Why exactly do we have $\pi^{-1}(H)=k \Bbb Z$? – MyNameIs Jul 22 '16 at 15:47
  • @MyNameIs $\pi^{-1}(H)=a\mathbb{Z}$ for a unique $a>0$; but $n/a=d$ because $\pi(\pi^{-1}(H))=\pi(a\mathbb{Z})=H$ has $d$ elements, so $a=k$. – egreg Jul 22 '16 at 15:49
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$G = \mathbb{Z}/n\mathbb{Z}$ is cyclic, so let $g$ be any generator.

Of course $\langle g^{n/d}\rangle$ is a subgroup of order $d$.

Let $a \in G$ be any order $d$ element. Then $a^d = 1$; in particular $a = g^t$ for some $t$ so $g^{td} = 1 \implies td \equiv 0 \pmod n$ so $t$ is a multiple of $n/d$. In particular, $t \in \langle g^{n/d} \rangle$ so $\langle t \rangle \leq \langle g^{n/d} \rangle$. They are both order $d$, hence equal.

Comments on your proof: in the first line, you use Lagrange's theorem. However I'm confused what you mean by $n/d \cdot g \in G/H$ is trivial. The elements of $G/H$ are cosets of the form $gH$ (or $a + H$ might be better notation here, since these are additive groups), not elements $g$.

MT_
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Every subgroup of $G$ is cyclic, so if $K$ is another subgroup of order $d$ it will generated by the "least integer" in $K.$ If this generator is $t,$ then the order of $t$ is $d=\frac{n}{gcd(t,n)}$ by the formula for the order of an element in a cyclic group of order $n.$ As $d=\frac{n}{gcd(n/d,d)}$ as well, we must have $gcd(t,n)=gcd(n/d,n).$ But, $gcd(n/d,n)=n/d,$ so $t$ is a multiple of $n/d.$ Therefore, $K=\langle t \rangle \subseteq H.$ But both $H$ and $K$ have $d$ elements, so they must be equal.

Chris Leary
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