The Sylow theorems do more than prove non-simplicity; they also help organise the early classification of finite groups. Small orders give a training ground for recognizing cyclic groups, direct products, dihedral groups, and quaternion behavior. In this lesson, the focus keyword is groups of small order. The aim is not to memorize a table blindly, but to understand why order, element orders, commutativity, and Sylow subgroups distinguish the main examples. :::definition[Classification by Isomorphism] Let $n$ be a positive integer. To classify groups of order $n$ means to list one representative from each isomorphism class of groups having order $n$. ::: For very small orders, the classification is short. There is one group of order $1$, one group of each prime order, two groups of order $4$, two groups of order $6$, five groups of order $8$, two groups of order $9$, and two groups of order $10$. The common representatives are $C_n$, direct products of cyclic groups, symmetric groups, dihedral groups, and the quaternion group $Q_8$. This preview compares small groups using invariants that are preserved by isomorphism. Choose a representative and read whether it is abelian, its largest element order, and how many elements of order $2$ it has. These quick checks distinguish many small groups without writing full multiplication tables. Use the order $8$ presets to compare $C_8$, $C_4\times C_2$, $C_2\times C_2\times C_2$, $D_4$, and $Q_8$. :::scientific-preview[Small Group Identifier Explorer] ::: Isomorphic groups have the same element-order data and the same commutativity behavior. Differences in these invariants are enough to separate many of the small groups listed in the notes. This calculator turns the comparison method into a diagnostic. Enter an order and a few observable invariants. The output suggests which listed examples fit those invariants and explains the distinguishing feature. It is designed for the small groups named in this lesson, not for a complete classification of every finite group. :::calculator[Small Group Distinguishing Calculator] ::: The calculator supports classification reasoning by contradiction: if two groups differ in any displayed invariant, they cannot be isomorphic. A match does not replace a proof, but it points to the right representative to investigate. :::theorem Every group of prime order is cyclic. ::: :::proof Given that $G$ is a group of prime order $p$. To prove that $G$ is cyclic. Choose $a\in G$ with $a\neq e$. By Lagrange's theorem, the order of $a$ divides $p$. Since $a\neq e$, we have $o(a)\neq 1$. Therefore $$\begin{equation*} o(a)=p. \end{equation*}$$ Thus $\langle a\rangle$ has $p$ elements. Since $|G|=p$, we get $$\begin{equation*} G=\langle a\rangle. \end{equation*}$$ Hence $G$ is cyclic. $\square$ ::: :::example The groups of order $8$ include $C_8$, $C_4\times C_2$, $C_2\times C_2\times C_2$, $D_4$, and $Q_8$. ::: :::solution The group $C_8$ has an element of order $8$. The group $C_4\times C_2$ has elements of order $4$, but no element of order $8$. The group $C_2\times C_2\times C_2$ has every nonidentity element of order $2$. These three abelian groups are distinguished by their element orders. The group $D_4$ is nonabelian. The group $Q_8$ is also nonabelian, but $Q_8$ has exactly one element of order $2$, while $D_4$ has several reflections of order $2$. Therefore $D_4$ and $Q_8$ are not isomorphic. Hence the five groups listed are pairwise nonisomorphic. $\square$ ::: :::solved-problem Classify the groups of order $14$ up to isomorphism. ::: :::solution Let $|G|=14=2\cdot 7$. By the classification of groups of order $2p$, where $p=7$, the group $G$ is either cyclic or dihedral. If $G$ is cyclic, then $$\begin{equation*} G\cong C_{14}. \end{equation*}$$ If $G$ is noncyclic, then $$\begin{equation*} G\cong D_7. \end{equation*}$$ Therefore, up to isomorphism, the groups of order $14$ are $$\begin{equation*} C_{14}\quad \text{and}\quad D_7. \end{equation*}$$ $\square$ ::: :::solved-problem Show that $C_{12}$, $C_2\times C_6$, $C_2\times S_3$, and $A_4$ are pairwise nonisomorphic. ::: :::solution The group $C_{12}$ is cyclic and has an element of order $12$. The group $C_2\times C_6$ is abelian but not cyclic; its largest element order is $6$. The group $C_2\times S_3$ is nonabelian because $S_3$ is nonabelian. The group $A_4$ is also nonabelian, so we need a finer distinction. In $A_4$, there are eight elements of order $3$ and three elements of order $2$. In $C_2\times S_3$, the element-order distribution differs because the extra $C_2$ factor changes the number of elements of order $2$ and $6$. Thus the four groups are distinguished by cyclicity, commutativity, maximum element order, and element counts. Hence they are pairwise nonisomorphic. $\square$ ::: :::solved-problem Determine the groups of order $70$ arising from the standard normal subgroup argument. ::: :::solution Let $|G|=70=2\cdot 5\cdot 7$. The number of Sylow $7$-subgroups satisfies $$\begin{align*} n_7 &\equiv 1 \pmod 7, \\ n_7 &\mid 10, \end{align*}$$ so $n_7=1$. The number of Sylow $5$-subgroups satisfies $$\begin{align*} n_5 &\equiv 1 \pmod 5, \\ n_5 &\mid 14, \end{align*}$$ so $n_5=1$. Thus $G$ has a normal cyclic subgroup of order $35$. Hence $G$ is a semidirect product $$\begin{equation*} C_{35}\rtimes C_2. \end{equation*}$$ The possible actions of $C_2$ on $C_{35}$ give the standard representatives $$\begin{equation*} C_{70},\quad D_5\times C_7,\quad C_5\times D_7,\quad D_{35}. \end{equation*}$$ Hence these four groups arise. $\square$ ::: :::exercise 1. List the groups of order $4$ up to isomorphism. 2. List the groups of order $6$ up to isomorphism. 3. Explain why $C_8$ is not isomorphic to $C_4\times C_2$. ::: :::answer 1. The groups of order $4$ are $C_4$ and $C_2\times C_2$. 2. The groups of order $6$ are $C_6$ and $S_3$, with $S_3\cong D_3$. 3. The group $C_8$ has an element of order $8$, while $C_4\times C_2$ has no element of order $8$. ::: :::faq[Frequently Asked Questions] Q: What is the fastest way to distinguish small groups? A: First compare abelian versus nonabelian, then compare element orders, centres, and numbers of elements of each order. Q: Why does Sylow theory appear in classification? A: Unique Sylow subgroups become normal, and normal subgroups often force direct or semidirect product decompositions. Q: Why is order $8$ richer than order $6$? A: There are several abelian and nonabelian structures of order $8$, including $D_4$ and $Q_8$. :::