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Section 10.1 Isomorphisms and Cayley's Theorem

Recall that two groups \((G, \cdot)\) and \((H, \circ)\) are isomorphic if there exists a one-to-one and onto map \(\phi : G \rightarrow H\) such that the group operation is preserved; that is,

\begin{equation*} \phi( a \cdot b) = \phi( a) \circ \phi( b) \end{equation*}

for all \(a\) and \(b\) in \(G\text{.}\) If \(G\) is isomorphic to \(H\text{,}\) we write \(G \cong H\text{.}\) The map \(\phi\) is called an isomorphism.

The main goal in group theory is to classify all groups; however, it makes sense to consider two groups to be the same if they are isomorphic. Hence, we can modify our goal of classifying all groups to classifying all groups up to isomorphism.

For example, you might wonder how many different cyclic groups there are of a given order. It turns out the answer is 1. For any given order, there is exactly one cyclic group (up to isomorphism) of that order: \(\Z_n\) is isomorphic to every cyclic group of order \(n\text{.}\) Later will will give a similar classification of all abelian groups of a particular type.

The other advantage in considering isomorphisms is that it provides us a way to represent groups in some consistent way. Let's see what this means.

Worksheet 10.1.1 Activity: Cayley Permutations

We have studied permutation groups, as well as groups that do not appear to be groups of permutations (such as \(\Z_n\text{,}\) \(\Z_n\times \Z_m\text{,}\) groups of functions or matrices, \(D_4\text{,}\) etc.). How distinct are these non-permutation groups from permutation groups?


Write the \(4\times 4\) group tables for \(\Z_4\) and \(U(8)\text{,}\) the group of units of \(\Z_8\) (numbers relatively prime to 8) under multiplication.

\begin{equation*} \begin{array}{c|cccc} + & 0 & 1 & 2 & 3 \\ \hline 0 & 0 & 1 & 2 & 3 \\ 1 & 1 & 2 & 3 & 0 \\ 2 & 2 & 3 & 0 & 1 \\ 3 & 3 & 0 & 1 & 2 \end{array} \qquad \end{equation*}

For each element in the groups above, we can see what adding or multiplying it by the other elements does to the other elements. For example, \(5 \in U(8)\) corresponds to this function: \(\lambda_5 = \begin{pmatrix}1 \amp 3 \amp 5 \amp 7\\ 5 \amp 7 \amp 1 \amp 3\end{pmatrix}\text{,}\) since \(5 \cdot 1 = 5 \text{,}\) \(5 \cdot 3 = 7\text{,}\) and so on.

For each element \(g\) in each group above, write down the corresponding function \(\lambda_g\text{.}\)


For \(Z_4\) we have

\begin{equation*} \lambda_0 = \begin{pmatrix}0 \amp 1 \amp 2 \amp 3 \\ 0 \amp 1 \amp 2 \amp 3\end{pmatrix} \end{equation*}
\begin{equation*} \lambda_1 = \begin{pmatrix}0 \amp 1 \amp 2 \amp 3 \\ 1 \amp 2 \amp 3 \amp 0\end{pmatrix} \end{equation*}
\begin{equation*} \lambda_2 = \begin{pmatrix}0 \amp 1 \amp 2 \amp 3 \\ 2 \amp 3 \amp 0 \amp 1\end{pmatrix} \end{equation*}
\begin{equation*} \lambda_3 = \begin{pmatrix}0 \amp 1 \amp 2 \amp 3 \\ 3 \amp 0 \amp 1 \amp 2\end{pmatrix} \end{equation*}

For \(U(8)\) we get

\begin{equation*} \lambda_1 = \begin{pmatrix}1 \amp 3 \amp 5 \amp 7\\ 1 \amp 3 \amp 5 \amp 7\end{pmatrix} \end{equation*}
\begin{equation*} \lambda_3 = \begin{pmatrix}1 \amp 3 \amp 5 \amp 7\\ 3 \amp 1 \amp 7 \amp 5\end{pmatrix} \end{equation*}
\begin{equation*} \lambda_5 = \begin{pmatrix}1 \amp 3 \amp 5 \amp 7\\ 5 \amp 7 \amp 1 \amp 3\end{pmatrix} \end{equation*}
\begin{equation*} \lambda_7 = \begin{pmatrix}1 \amp 3 \amp 5 \amp 7\\ 7 \amp 5 \amp 3 \amp 1\end{pmatrix} \end{equation*}

Notice that we are getting the rows in the group table as the outputs of the function. This makes sense, right?


What happens when you compose two functions \(\lambda_g\) and \(\lambda_h\) for \(g\) and \(h\) in a group? Try this with a few examples you have above. What do you notice about the function you get?


For example, in the case of \(\Z_4\text{,}\) we have

\begin{equation*} \lambda_2\lambda_1 = \begin{pmatrix}0 \amp 1 \amp 2 \amp 3\\ 2 \amp 3 \amp 0 \amp 1 \end{pmatrix}\begin{pmatrix}0 \amp 1 \amp 2 \amp 3\\ 1 \amp 2 \amp 3 \amp 0 \end{pmatrix} = \begin{pmatrix}0 \amp 1 \amp 2 \amp 3\\ 3 \amp 0 \amp 1 \amp 2 \end{pmatrix} = \lambda_3 \text{.} \end{equation*}

In \(U(8)\text{,}\) we have \(\lambda_3\lambda_7 = \lambda_5\text{.}\) But notice that \(5 = 3 \cdot 7\) in \(U(8)\text{.}\) In other words, the \(\lambda_g\) are acting just like \(g\text{.}\)


Consider the set \(\overline{G} = \{\lambda_g \st g \in G\}\text{.}\) Since each \(\lambda_g\) is a permutation of the elements of \(G\text{,}\) each of these will be a subset of \(S_n\) where \(n = |G|\) (in our case, \(n=4\)). Is \(\overline{G}\) a subgroup of \(S_4\) in both our cases? Will it be a subgroup in general?


Instead of using the elements of \(G\) themselves, if we put them in some order (a first one, second one, etc), then we can associate numbers \(1,2,3,\ldots,n\) to the \(n\) elements of \(G\text{.}\) The permutations \(\lambda_g\) are then exactly elements of \(S_n\text{.}\)

What we found in the previous question is that the \(\lambda_g\) act just like the elements \(g\) they come from. In fact, we can prove that \(G \cong \overline{G}\text{:}\) the set of permutations \(\overline{G}\) is a subgroup of \(S_n\text{,}\) in fact, a subgroup isomorphic to \(G\text{.}\)

Subsection 10.1.2 Cayley's Theorem

Cayley proved that if \(G\) is a group, it is isomorphic to a group of permutations on some set; hence, every group is a permutation group. Cayley's Theorem is what we call a representation theorem. The aim of representation theory is to find an isomorphism of some group \(G\) that we wish to study into a group that we know a great deal about, such as a group of permutations or matrices.

Example 10.1.

Consider the group \({\mathbb Z}_3\text{.}\) The Cayley table for \({\mathbb Z}_3\) is as follows.

\begin{equation*} \begin{array}{c|ccc} + & 0 & 1 & 2 \\ \hline 0 & 0 & 1 & 2 \\ 1 & 1 & 2 & 0 \\ 2 & 2 & 0 & 1 \end{array} \end{equation*}

The addition table of \({\mathbb Z}_3\) suggests that it is the same as the permutation group \(G = \{ (0), (0 1 2), (0 2 1) \}\text{.}\) The isomorphism here is

\begin{align*} 0 & \mapsto \begin{pmatrix} 0 & 1 & 2 \\ 0 & 1 & 2 \end{pmatrix} = (0)\\ 1 & \mapsto \begin{pmatrix} 0 & 1 & 2 \\ 1 & 2 & 0 \end{pmatrix} = (0 1 2)\\ 2 & \mapsto \begin{pmatrix} 0 & 1 & 2 \\ 2 & 0 & 1 \end{pmatrix} = (0 2 1)\text{.} \end{align*}

You will be guided through a proof of this theorem in the homework.

The isomorphism \(g \mapsto \lambda_g\) is known as the left regular representation of \(G\) (“left” because we get the permutation by multiplying each element in the group on the left; there are also right regular representations, but we will not consider these).

Subsection 10.1.3 Historical Note

Arthur Cayley was born in England in 1821, though he spent much of the first part of his life in Russia, where his father was a merchant. Cayley was educated at Cambridge, where he took the first Smith's Prize in mathematics. A lawyer for much of his adult life, he wrote several papers in his early twenties before entering the legal profession at the age of 25. While practicing law he continued his mathematical research, writing more than 300 papers during this period of his life. These included some of his best work. In 1863 he left law to become a professor at Cambridge. Cayley wrote more than 900 papers in fields such as group theory, geometry, and linear algebra. His legal knowledge was very valuable to Cambridge; he participated in the writing of many of the university's statutes. Cayley was also one of the people responsible for the admission of women to Cambridge.

Exercises 10.1.4 Collected Homework


Let's prove Cayley's theorem. Here is the overall structure, including the “big idea”. After this, you are asked to provide the details.


Let \(G\) be a group. We will define a group of permuations \(\overline{G}\) that is isomorphic to \(G\text{.}\) For any \(g \in G\text{,}\) define a function \(\lambda_g : G \rightarrow G\) by \(\lambda_g(a) = ga\text{.}\) [Show \(\lambda_g\) is a permutation of \(G\text{,}\) i.e., a bijection.]

Now we are ready to define our group \(\overline{G}\text{.}\) Let

\begin{equation*} \overline{G} = \{ \lambda_g \st g \in G \}\text{.} \end{equation*}

We must show that \(\overline{G}\) is a group (where the operation is composition of functions). [Show that \(\overline{G}\) is subgroup of \(S_G\) by showing it contains the identity, is closed under the operation, and is closed under inverses.]

We can define an isomorphism from \(G\) to \(\overline{G}\) by \(\phi : g \mapsto \lambda_g\text{.}\) We must prove this is really an isomorphism. [Show that \(\phi\) is a bijection and satisfies the homomorphism property.]

Now fill in the gaps:

  1. Prove that \(\lambda_g\) is injective (one-to-one).

  2. Prove that \(\lambda_g\) is surjective (onto).

  3. Prove that \(\overline{G}\) contains the identity.

  4. Prove that \(\overline{G}\) is closed under the operation (composition).

  5. Prove that \(\overline{G}\) is closed under inverses.

  6. Prove that \(\phi\) is injective.

  7. Prove that \(\phi\) is surjective.

  8. Prove that \(\phi\) satisfies the homomorphism property.


To prove a function \(f:A \to B\) is injective, suppose \(f(a) = f(b)\) and deduce that \(a = b\text{.}\) To prove that the function is surjective, assume \(b \in B\) and find some \(a \in A\) such that \(f(a) = b\text{.}\)