Cyclic Groups and Direct Products

For $(g,h),(g',h'),(g'',h'')\in G\times H$, associativity in each factor gives $((g,h)(g',h'))(g'',h'')=(gg'g'',hh'h'')=(g,h)((g',h')(g'',h''))$; thus the componentwise operation is associative.

For every $(g,h)\in G\times H$, $(e_G,e_H)(g,h)=(g,h)=(g,h)(e_G,e_H)$; thus $(e_G,e_H)$ is a two-sided identity.

For every $(g,h)\in G\times H$, $(g,h)(g^{-1},h^{-1})=(e_G,e_H)=(g^{-1},h^{-1})(g,h)$; so $(g^{-1},h^{-1})$ is its inverse.

Steps 1.1–1.3 verify the group axioms for $G\times H$.

For pairs $(g,h),(g',h')$, $\pi_G((g,h)(g',h'))=gg'=\pi_G(g,h)\pi_G(g',h')$; the same coordinatewise calculation holds for $\pi_H$, so both projections are homomorphisms from the group in step 2.1.

The stated identity, inverse formula, and coordinate homomorphisms follow.

The carrier of the direct-product group is the Cartesian product of the finite carriers $G$ and $H$.

By the finite product rule, $G\times H$ is finite and $|G\times H|_{\mathrm{set}}=|G|_{\mathrm{set}}|H|_{\mathrm{set}}$.

Reading these three cardinalities as group orders gives $|G\times H|=|G|\,|H|$.

For every natural $k$, $(g,h)^k=(g^k,h^k)$: it holds at $k=0$, and the successor step follows by componentwise multiplication.

Let $\ell$ be the natural from [L3]. Since $\iota(m)\mid\iota(\ell)$ and $\iota(n)\mid\iota(\ell)$, [L2] and step 1.1 give $(g,h)^\ell=(e_G,e_H)$.

If $(g,h)^k=(e_G,e_H)$ for a positive natural $k$, then step 1.1 gives $g^k=e_G$ and $h^k=e_H$. Hence $\iota(m)\mid\iota(k)$ and $\iota(n)\mid\iota(k)$.

By [L3], the two divisibilities of step 2.2 imply $\iota(\ell)\mid\iota(k)$. As $\ell,k\ge1$, this forces $\ell\le k$: an integer quotient $q$ with $\iota(k)=q\iota(\ell)$ is positive and hence at least $1$. Thus $\ell$ is the least positive exponent sending $(g,h)$ to the identity.

The definition of element order gives $\operatorname{ord}(g,h)=\ell$. Applying $\iota$ and using [L3] gives the displayed equality.

Suppose first that $g$ has infinite order. The map $\phi:\mathbb Z\to G$, $r\mapsto g^r$, is a homomorphism by the power law in [L3]. If $g^r=g^s$, then $g^{r-s}=g^r(g^s)^{-1}=e$ by [L3], so [L2] forces $r=s$; it is surjective by [L1].

Suppose instead that $g$ has finite order $n$. Then $n\ge1$ by the definition of element order, and $[r]\mapsto g^r$ defines a map $\bar\phi:\mathbb Z/n\to G$ because $[r]=[s]$ means $n\mid(r-s)$, so [L2] gives $g^{r-s}=e$ and [L3] gives $g^r=g^{r-s}g^s=g^s$.

Hence in the infinite-order case $\phi$ is an isomorphism $\mathbb Z\cong G$.

The map $\bar\phi$ is a homomorphism by the power law, is injective because $g^r=g^s$ gives $g^{r-s}=g^r(g^s)^{-1}=e$ by [L3] and then [L2] gives $n\mid(r-s)$, and is surjective by [L1].

Thus $\bar\phi$ is an isomorphism $(\mathbb Z/n,+)\cong G$ in the finite-order case.

Steps 2.1 and 3.1 give the asserted classification according as the order of $g$ is infinite or finite.

If $H=\{e\}$, then $H=\langle e\rangle$ and is cyclic. Assume henceforth that $H\ne\{e\}$.

The set $S:=\{n\in\mathbb N:n\ge1\text{ and }g^n\in H\}$ is nonempty: choose $h\in H\setminus\{e\}$ and write $h=g^k$ by [L2]. Since $g^0=e$, $k\ne0$; by [L6], either $k>0$ or $-k>0$. In the first case use $h$, and in the second use $h^{-1}=g^{-k}$.

Let $d$ be the least element of $S$, supplied by [L4]. Then $g^d\in H$, so $\langle g^d\rangle\subseteq H$ by the smallest-subgroup property in [L2].

For $h\in H$, write $h=g^k$ by [L2] and write $k=qd+r$ with $0\le r<d$ by [L5]. Then $g^r=h(g^d)^{-q}\in H$.

If $r>0$ in step 3.1, then $r\in S$ contradicts the minimality of $d$; hence $r=0$, so $h=(g^d)^q\in\langle g^d\rangle$.

Thus $H\subseteq\langle g^d\rangle$ in the nontrivial case, and step 2.1 gives equality; together with step 1.1 this proves every subgroup of $G$ is cyclic.