FORM GR1268 Question 12

Question:
For which integers \(n\) such that \(3\le n\le 11\) is there only one group of order \(n\) (up to isomorphism)?

(A) For no such integer \(n\)
(B) For \(3,5,7,\) and \(11\) only
(C) For \(3,5,7,9,\) and \(11\) only
(D) For \(4,6,8,\) and \(10\) only
(E) For all such integers \(n\)

Answer:
(B)

Answer Key:
There is only one group of order \(3\), namely, \(C_3\), a cyclic group of order \(3\).

To see this, consider a set \(\{1,a,b\}\).
A partially filled multiplication table looks like this:

\[\begin{array}{|c|c|c|c|}
\hline
\times &\textbf{1} & \textbf{a} & \textbf{b}\\ \hline
\textbf{1} & 1 & a & b\\ \hline
\textbf{a} & a & ? &\\ \hline
\textbf{b} & b & & \\ \hline
\end{array}\]

Since no row or column can have any element more than once,
it must be that \(a\times a = 1\) or \(a\times a = b\).
If \(a\times a = 1\),then \(a\times b = b\), which implies \(a=1\), which is not the case.
Thus, it must be that \(a\times a = b\), which leaves us with \(a\times b = 1\).

\[\begin{array}{|c|c|c|c|}
\hline
\times &\textbf{1} & \textbf{a} & \textbf{b}\\ \hline
\textbf{1} & 1 & a & b\\ \hline
\textbf{a} & a & b & 1\\ \hline
\textbf{b} & b & & \\ \hline
\end{array}\]

There is only one way to complete the last row:
\[\begin{array}{|c|c|c|c|}
\hline
\times &\textbf{1} & \textbf{a} & \textbf{b}\\ \hline
\textbf{1} & 1 & a & b\\ \hline
\textbf{a} & a & b & 1\\ \hline
\textbf{b} & b & 1 & a\\ \hline
\end{array}\]

This is a cyclic group of order \(3\).
Note that it is isomorphic to \(\mathbb Z/3\mathbb Z\).

So, the correct answer is (A), (B), or (E).

Is there a non-cyclic group of order \(9\)?

The answer is yes.
Consider two cyclic groups \(\{1,a,a^2\}\) and \(\{1,b,b^2\}\).
The Cartesian product forms a new group:
\[\{1,a,a^2\}\times\{1,b,b^2\}=\{1, b, b^2, a, ab, ab^2, a^2, a^2b, a^2b^2\}.\]
The operation is inherited from the original cyclic groups.
For example, \(ab\times a^2b=a^3b^2=1b^2=b^2\).
Convince yourself this set is closed.

Is it a cyclic group?
We know \(1\) is not a generator.
Neither \(a\) nor \(a^2\) is a generator because it cannot generate \(b\).
Neither \(b\) nor \(b^2\) is a generator because it cannot generate \(a\).
Neither \(ab\) nor \(a^2b^2\) is a generator because it cannot generate \(a\).
Neither \(a^2b\) nor \(ab^2\) is a generator because it cannot generate \(a\).
Thus, the group is non-cyclic.