Isomorphisms (AQA A-Level Further Maths): Revision Notes

Worked Example: Proving Isomorphism Between Complex Numbers Group and Group G

Question: A group $G$ under the binary operation $\bullet$ has the Cayley table:

$\bullet$	$p$	$q$	$r$	$s$
$p$	$p$	$q$	$r$	$s$
$q$	$q$	$p$	$s$	$r$
$r$	$r$	$s$	$q$	$p$
$s$	$s$	$r$	$p$	$q$

A second group $H = (\{i, -i, 1, -1\}, \times)$ consists of the set of complex numbers with multiplication as the operation.

Show that $H$ is isomorphic to $G$ and state the corresponding elements in each group.

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Solution:

First, construct the Cayley table for $H$:

$\times$	$i$	$-i$	$1$	$-1$
$i$	$-1$	$1$	$i$	$-i$
$-i$	$1$	$-1$	$-i$	$i$
$1$	$i$	$-i$	$1$	$-1$
$-1$	$-i$	$i$	$-1$	$1$

Now identify key elements:

Identity element for $H$: The identity is 1 since $1 \times a = a \times 1 = a$ for all $a$.

Self-inverse elements in $H$: We need elements where $a \times a = 1$:

• $1 \times 1 = 1$ ✓
• $(-1) \times (-1) = 1$ ✓

So 1 and -1 are self-inverse.

From group $G$, by examining the Cayley table, we can see that p is the identity (first row and column are unchanged), and both $p$ and $q$ are self-inverse.

Establishing correspondences:

• Since $1$ is the identity in $H$ and $p$ is the identity in $G$: $1 \leftrightarrow p$
• Since $-1$ is self-inverse in $H$ and $q$ is self-inverse in $G$: $-1 \leftrightarrow q$
• For non-self-inverse elements, we need to check which pairing works: $i \leftrightarrow r$ and $-i \leftrightarrow s$

Now reorder the Cayley table for $H$ with columns and rows in the order $1, -1, i, -i$:

$\times$	$1$	$-1$	$i$	$-i$
$1$	$1$	$-1$	$i$	$-i$
$-1$	$-1$	$1$	$-i$	$i$
$i$	$i$	$-i$	$-1$	$1$
$-i$	$-i$	$i$	$1$	$-1$

By comparing this reordered table with the Cayley table for $G$, we can see that the pattern of entries is identical. Therefore, the groups are isomorphic.

The corresponding elements are: $1 \leftrightarrow p, \quad -1 \leftrightarrow q, \quad i \leftrightarrow r, \quad -i \leftrightarrow s$

Hence $H \cong G$.

Note: The correspondence is not unique. Interchanging $i$ and $-i$ (which would correspond to interchanging $r$ and $s$) would also give the same pattern of entries in the Cayley table.

Worked Example: Systematic Proof Using the Five-Step Method

Question: Groups $G$ and $H$ have Cayley tables as shown below. Show that $H$ is isomorphic to $G$.

Group $G$:

$\bullet$	$a$	$b$	$c$	$d$	$e$	$f$
$a$	$a$	$b$	$c$	$d$	$e$	$f$
$b$	$b$	$c$	$a$	$f$	$d$	$e$
$c$	$c$	$a$	$b$	$e$	$f$	$d$
$d$	$d$	$e$	$f$	$a$	$b$	$c$
$e$	$e$	$f$	$d$	$c$	$a$	$b$
$f$	$f$	$d$	$e$	$b$	$c$	$a$

Group $H$:

$\star$	$P$	$Q$	$R$	$S$	$T$	$U$
$P$	$S$	$U$	$T$	$P$	$R$	$Q$
$Q$	$T$	$R$	$S$	$Q$	$U$	$P$
$R$	$U$	$S$	$Q$	$R$	$P$	$T$
$S$	$P$	$Q$	$R$	$S$	$T$	$U$
$T$	$Q$	$P$	$U$	$T$	$S$	$R$
$U$	$R$	$T$	$P$	$U$	$Q$	$S$

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Solution:

Following the systematic five-step approach:

Step 1: Identify the identity element in each group

In group $G$, the identity element is a because the first row shows $a \bullet x = x$ for all elements.

In group $H$, the identity element is S because the row for $S$ shows $S \star x = x$ for all elements.

Step 2: Find the self-inverse elements in each group

In group $G$, checking the diagonal:

• $d \bullet d = a$ ✓
• $e \bullet e = a$ ✓
• $f \bullet f = a$ ✓

So the self-inverse elements in G are d, e, f (plus the identity $a$).

In group $H$, checking the diagonal:

• $P \star P = S$ ✓
• $T \star T = S$ ✓
• $U \star U = S$ ✓

So the self-inverse elements in H are P, T, U (plus the identity $S$).

Step 3: Identify elements that are not self-inverse and choose an arbitrary mapping

In group $G$: $b$ and $c$ are not self-inverse.

In group $H$: $Q$ and $R$ are not self-inverse.

Choose an arbitrary mapping, say $b \leftrightarrow Q$. This means that $c \leftrightarrow R$.

Step 4: Choose an arbitrary pairing of the self-inverse elements

Let's pair: $d \leftrightarrow P$, $e \leftrightarrow U$, $f \leftrightarrow T$.

Our complete mapping is: $a \leftrightarrow S, \quad b \leftrightarrow Q, \quad c \leftrightarrow R, \quad d \leftrightarrow P, \quad e \leftrightarrow U, \quad f \leftrightarrow T$

Step 5: Rewrite the Cayley table for $H$ using this mapping

Reordering $H$'s table with columns and rows in the order $S, Q, R, P, U, T$:

$\star$	$S$	$Q$	$R$	$P$	$U$	$T$
$S$	$S$	$Q$	$R$	$P$	$U$	$T$
$Q$	$Q$	$R$	$S$	$T$	$P$	$U$
$R$	$R$	$S$	$Q$	$U$	$T$	$P$
$P$	$P$	$U$	$T$	$S$	$Q$	$R$
$U$	$U$	$T$	$P$	$R$	$S$	$Q$
$T$	$T$	$P$	$U$	$Q$	$R$	$S$

Comparing this redrawn Cayley table with the Cayley table for group $G$, we can see that the pattern of entries matches exactly when we use the correspondence established above.

Therefore, $H \cong G$.

Worked Example: Proving Groups Are NOT Isomorphic

Question: Group $S$ has a Cayley table as shown. Show that $S$ is not isomorphic to groups of order 4.

$\circ$	$1$	$2$	$3$	$4$	$5$	$6$
$1$	$1$	$2$	$3$	$4$	$5$	$6$
$2$	$2$	$4$	$6$	$1$	$3$	$5$
$3$	$3$	$6$	$2$	$5$	$1$	$4$
$4$	$4$	$1$	$5$	$2$	$6$	$3$
$5$	$5$	$3$	$1$	$6$	$4$	$2$
$6$	$6$	$5$	$4$	$3$	$2$	$1$

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Solution:

First, identify the identity element and self-inverse elements in group $S$.

Identity element: From the table, 1 is the identity since $1 \circ a = a$ for all elements.

Self-inverse elements: Check which elements satisfy $a \circ a = 1$:

• $1 \circ 1 = 1$ ✓
• $6 \circ 6 = 1$ ✓

So group $S$ has only two self-inverse elements: 1 and 6.

Checking periods of elements in $S$:

For element $3$:

• $3 \circ 3 = 2$
• $3^3 = 2 \circ 3 = 6$
• $3^4 = 6 \circ 3 = 4$
• $3^5 = 4 \circ 3 = 5$
• $3^6 = 5 \circ 3 = 1$

So element 3 has period 6.

Similarly, element $5$ has period $6$ (you can verify this by repeated operations).

However, in groups of order 4, no element can have period greater than 4 by Lagrange's theorem.

Therefore, S is not isomorphic to any group of order 4 because $S$ has elements of period 6, whereas groups of order 4 cannot have elements with period greater than 4.

The key point is: different structural properties mean the groups cannot be isomorphic.