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Let $G$ be a group of order $n$. Prove that $G$ has at most $n$ different cyclic subgroups.

I know that the generators from the group form the cyclic subgroup. The number of generators, at most, is the number of elements from the group, ($n$). How to go about proving this?

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Basically, the argument is "Any cyclic subgroup is generated by at least one element; therefore there can't be any more cyclic subgroups than there are elements of $G$."

To be more formal, let $X$ be the set of cyclic subgroups of $G$, i.e., $$X=\{H\subseteq G:H = \langle g\rangle \text{ for some }g\in G\}$$ Let $f:G\to X$ be the function that takes an element $g\in G$, and outputs $\langle g\rangle\in X$, the cyclic subgroup of $G$ generated by that element $g$. In symbols, this would be written $f(g)=\langle g\rangle$.

Any cyclic subgroup $H$ of $G$ is in fact generated by some element of $G$ (by definition). That is, any element of $X$ equals $f(g)$ for some $g\in G$. This simply says that $f:G\to X$ is surjective, and therefore the cardinality of $X$ is less than or equal to the cardinality of $G$.

Note that $f$ is usually not injective; that is, many elements will generate the same cyclic subgroup. For example, in $G=\mathbb{Z}_{6}$, we have the cyclic subgroups $$X=\{\{0\},\{0,3\},\{0,2,4\},\{0,1,2,3,4,5\}\}$$ and the function $f:G\to X$ acts like this: $$\begin{align*} f(0)=\langle 0\rangle &= \{0\}\\ f(1)=\langle 1\rangle &= \{0,1,2,3,4,5\}\\ f(2)=\langle 2\rangle &= \{0,2,4\}\\ f(3)=\langle 3\rangle &= \{0,3\}\\ f(4)=\langle 4\rangle &= \{0,2,4\}\\ f(5)=\langle 5\rangle &= \{0,1,2,3,4,5\} \end{align*}$$

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Telling a cyclic group is equivalent to telling a generator of it; since $G$ has order $n$, the $n$ elements will give at most $n$ cyclic groups.

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