# Proof: any element of a group of prime order generates the group

The following self-contained post clarifies the proof of theorem 4, chapter 13, p. 129, of the book “A book of abstract algebra” by Charles C. Pinter.

Let $G$ be a group of prime order $|G|=p$. It will be shown that for any $a\in G$ with $a\ne e$ it holds that $\langle a\rangle=G,$ where $\langle a\rangle$ is the set generated by $a$ and $e$ is the identity element of $G$.

Since the order of group $G$ is $|G|=p<\infty$, the order $\mbox{ord}(a)$ of element $a$ is also finite, i.e. $\mbox{ord}(a)=n<\infty$ for some positive integer $n$.

Note that the set $\langle a\rangle=\left\{a^0,a^1,\ldots,a^{n-1}\right\}$ generated by $a$ is a subgroup of $G$, since $\langle a\rangle$ is closed with respect to multiplication and with respect to inverses. Indeed, for any $a^i,a^j\in\langle a\rangle$, the division algorithm of integers ensures that there exist integers $q,r$ with $0\le r such that $a^i a^j = a^{i+j}=a^{nq+r}=(a^n)^q a^r=e a^r=a^r\in \langle a \rangle$. Moreover, the inverse of any $a^i\in \langle a \rangle$ is $(a^i)^{-1}=a^{n-i}\in \langle a \rangle$.

The order $|\langle a \rangle|$ of the cyclic subgroup $\langle a \rangle$ of $G$ is equal to the order of generator $a$, that is $|\langle a \rangle|=\mbox{ord}(a)=n$.

By Lagrange’s theorem, $n$ is a factor of $p$. Since $p$ is a prime, $n=1$ or $n=p$. It follows from $a\ne e$ that $n\ne 1$, so $n=p$, which means that $|\langle a \rangle|=|G|$.

$\langle a \rangle\subseteq G$ and $|\langle a \rangle|=|G|\Rightarrow\langle a \rangle=G$.

# Clarification on the proof of isomorphism of finite cyclic groups

The following self-contained post clarifies the proof of theorem 1, chapter 11, p. 113-114, of the book “A book of abstract algebra” by Charles C. Pinter.

In order to show that any two cyclic groups of finite order $n$ are isomorphic to each other, it suffices to show that every cyclic group $\langle a\rangle=\left\{a^0,a^1,a^2,\ldots,a^{n-1}\right\}$ of finite order $n$ is isomorphic to the group $\mathbb{Z}_n$ of integers modulo $n$ with the operation of addition modulo $n$, where $\mathbb{Z}_n=\left\{0,1,2,\ldots,n-1\right\}$.

Towards this end, it can be shown that the function $f:\mathbb{Z}_n\rightarrow\langle a\rangle$ defined as $f(i)=a^i$ is a group isomorphism. Recalling the definition of group isomorphism, it suffices to show that $f$ is a bijective homomorphism. It is straightforward to check that $f$ is a bijective function. The fact that $f$ is a homomorphism is proven also easily using laws of exponents for groups, as follows: $f(i+j)=a^{i+j}=a^i a^j=f(i)f(j)$.

Proving the homomorphic property of $f$ as above works well if $f$ is taken to be $f:\mathbb{Z}\rightarrow\langle a\rangle,~f(i)=a^i,$ that is from the additive group of integers to the cyclic group $\langle a\rangle=\left\{\ldots,a^{-2},a^{-1},a^0,a^1,a^2,\ldots\right\}$ of infinite order generated by $a$. However, in the case of $f:\mathbb{Z}_n\rightarrow\langle a\rangle$ with $\langle a\rangle=\left\{a^0,a^1,a^2,\ldots,a^{n-1}\right\}$, there is a slight oversight; the addition implied by $f(i+j)$ is addition modulo $n$.

Let’s denote the addition modulo $n$ by $\oplus$. So, now the proof of homomorphic property takes the form $f(i\oplus j)=a^{i\oplus j}=a^i a^j=f(i)f(j)$. The identity$a^{i+j}=a^i a^j$ is not the same as$a^{i\oplus j}=a^i a^j$; the former is one of the common exponent laws for groups, while the latter is not.

What needs to be shown is that the exponents of addition and of addition modulo $n$ coincide, i.e. $a^{i+j}=a^{i\oplus j}$. The proof of this identity is simple, yet it is useful to be aware of such potential oversight.

According to the division algorithm for integers, there exist unique integers $q$ and $r$ such that $i+j = nq+r$ and $0\le r < n$. Thus, $a^{i+j}=a^{nq+r}=(a^n)^q a^r$. Since the cyclic group $\langle a\rangle$ is of order $n$, it holds that $a^n=e$, where $e$ is the identity element of $\langle a\rangle$. So, $a^{i+j}=a^r$.

Notice that $r$ coincides with the result of the addition modulo $n$, that is $r=i\oplus j$. It follows that $a^{i+j}=a^{i\oplus j}$, which completes the proof.