Prove that if $k\mid m,$ then $Z_m$ has a subgroup of order $k.$

Question

Prove that if $k\mid m$, then $Z_m$ has a subgroup of order $k.$

Ok, so this doesn't look like too hard of a problem. So do I just show that it is closed under multiplication and inverse? I just don't know how I would do that with a problem like this. Can anyone help me get started please?

Answer 1

Suppose that $g$ generates $C_m(\simeq {\bf Z}_m)$, $k\mid m$. Consider $g'=g^{m/k}$. What is the order of $\langle g'\rangle$?

Answer 2

If $k \mid m$, then by definition, there exists $n \in \mathbb{Z}/m\,\mathbb{Z}$ such that

$$ m = kn$$

Consider this $k$-element subset of $\mathbb{Z}/m\,\mathbb{Z}$ (generated by $n$):

$$\left\{0, \,n,\, 2n,\, 3n,\, \ldots \,(k-1) n\right\}$$

This set, with the operation of addition as in $\mathbb{Z}/m\,\mathbb{Z}$, is closed, associative, has inverses, and includes the additive identity (zero).

Thus, we have constructed a subgroup of order $k$.

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This proof applies to the integers modulo $m$, but it works in exactly the same way for the cyclic group of order $m$.

Answer 3

Since $k|m$ then $m=kd$ for $d\in \mathbb{Z}$

Claim: $<\frac{m}{k}$> is the subgroup we desire.

Hint: Remember this group has operation addition.