Fundamental theorem of cyclic groups

In abstract algebra, the fundamental theorem of cyclic groups states that if $G,$ is a cyclic group of order $n,$ then every subgroup of $G,$ is cyclic. Moreover, the order of any subgroup of $G,$ is a divisor of $n,$ and for each positive divisor $k,$ of $n,$ the group $G,$ has at most one subgroup of order $k,$.

Proof

Let $G\; =\; langle\; a\; angle,$ be a cyclic group for some $a\; in\; G$ and with identity $e,$ and order $n,$, and let $H,$ be a subgroup of $G,$.We will now show that $H,$ is cyclic. If $H\; =\; \{\; e\; \},$ then $H\; =\; langle\; e\; angle,$. If $H\; eq\; \{\; e\; \},$ then since $G,$ is cyclic every element in $H,$ is of the form $a^t,$, where $t,$ is a positive integer. Let $m,$ be the least positive integer such that $a^m\; in\; H$.

We will now show that $H\; =\; langle\{a^m\}\; angle,$. It follows immediately from the closure property that $langle\; \{a^m\}\; angle\; subseteq\; H$.

To show that $H\; subseteq\; langle\; \{a^m\}\; angle$ we let $b\; in\; H$. Since $b\; in\; G$ we have that $b\; =\; a^k,$ for some positive integer $k,$. By the division algorithm, $k\; =\; mq\; +\; r,$ with , and so $a^k\; =\; a^\{mq\; +\; r\}\; =\; a^\{mq\}a^r,$, which yields $a^r\; =\; a^\{-mq\}a^k,$. Now since $a^k\; in\; H$ and $a^\{-mq\}\; =\; (a^\{m\})^\{-q\}\; in\; H$, it follows from closure that $a^r\; in\; H$. But $m,$ is the least integer such that $a^m\; in\; H$ and , which means that $r\; =\; 0,$ and so $b\; =\; a^k\; =\; a^\{mq\}\; =\; (a^m)^q\; in\; langle\; a^m\; angle$. Thus $H\; subseteq\; langle\; \{a^m\}\; angle$.

Since $H\; subseteq\; langle\; a^m\; angle$ and $langle\; \{a^m\}\; angle\; subseteq\; H$ it follows that $H\; =\; langle\; a^m\; angle\; ,$ and so $H,$ is cyclic.

We will now show that the order of any subgroup of $G,$ is a divisor of $n,$. Let $H,$ be any subgroup of $G,$. We have already shown that $H\; =\; langle\; a^m\; angle,$, where m is the least positive integer such that $a^m\; in\; H$. Since $e\; =\; a^n\; =\; a^m,$ it follows that $n\; =\; mq,$ for some integer $q,$. Thus $m\; |\; n,$.

We will now prove the last part of the theorem. Let $k,$ be any positive divisor of $n,$. We will show that $langle\; a^\{n/k\}\; angle,$ is the one and only subgroup of $langle\; a\; angle,$ of order $k,$. Note that $langle\; a^\{n/k\}\; angle,$ has order $\{nover\{gcd(n,\; \{n\; over\; \{k)\; =\; \{n\; over\; \{n\; over\; k\; =\; k,$. Let $H,$ be any subgroup of $langle\; a\; angle,$ with order $k,$. We know that $H\; =\; langle\; a^m\; angle,$, where $m,$ is a divisor of $n,$. So $m\; =\; operatorname\{gcd\}(n,\; m),$ and $k\; =\; |\; langle\; a\; ^m\; angle\; |\; =\; |\; a\; ^m\; |\; =\; |a^\{operatorname\{gcd\}(n,\; m)\}|\; =\; \{n\; over\; \{operatorname\{gcd\}(n,\; m)\; =\; \{n\; over\; m\},$. Consequently $m\; =\; \{n\; over\; k\},$ and so $H\; =\; langle\; a^\{n\; over\; k\}\; angle,$, and thus the theorem is proved.

Alternate proof

Let $G\; =\; langle\; a\; angle$ be a cyclic group, and let $H$ be a subgroup of $G$. Define a morphism $varphi:\; mathbb\{Z\}\; ightarrow\; G$ by $varphi(n)\; =\; a^n$. Since $G$ is cyclic generated by $a$, $varphi$ is surjective. Let $K\; =\; varphi^\{-1\}(H)\; subseteq\; mathbb\{Z\}$. $K$ is a subgroup of $mathbb\{Z\}$. Since $varphi$ is surjective, the restriction of $varphi$ to $K$ defines a surjective morphism from $K$ onto $H$, and therefore $H$ is isomorphic to a quotient of $K$. Since $K$ is a subgroup of $mathbb\{Z\}$, $K$ is $nmathbb\{Z\}$ for some integer $n$. If $n\; =\; 0$, then $K\; =\; \{0\}$, hence $H\; =\; \{0\}$, which is cyclic. Otherwise, $K$ is isomorphic to $mathbb\{Z\}$. Therefore $H$ is isomorphic to a quotient of $mathbb\{Z\}$, and is necessarily cyclic.

Converse

The following statements are equivalent.
* A group G of order $n,$ is cyclic.
* For every divisor $d,$ of $n,$ a group G has exactly one subgroup of order $d,$.
* For every divisor $d,$ of $n,$ a group G has at most one subgroup of order $d,$.

ee also

*Cyclic group