DG12 - Isomorphism of finite groups | AQA A-Level Further Mathematics

Different-looking groups can have exactly the same operation structure. You will construct bijections that preserve the operation, write an isomorphism using the correct symbol, and rule out proposed matches using element periods or commutativity. Equal group orders alone are not enough.

Bijections that preserve the operation

Do the four rotations of a square behave like the non-zero residues under multiplication modulo $5$ ? Their elements look different, but we can try matching each rotation with a residue so that combining either pair of labels gives matching answers.

First consider a smaller match. Addition modulo $2$ has elements $0,1$ ; multiplication on $\{1,-1\}$ has identity $1$ . Match $0\mapsto1$ and $1\mapsto-1$ . Then all four possible products correspond:

Add modulo $2$ first, then map	Map first, then multiply
$0+_2 0=0\mapsto1$	$1\times1=1$
$0+_2 1=1\mapsto-1$	$1\times(-1)=-1$
$1+_2 0=1\mapsto-1$	$(-1)\times1=-1$
$1+_2 1=0\mapsto1$	$(-1)\times(-1)=1$

The elements are renamed, and the rule of combination survives the renaming.

Group isomorphism

For groups $G$ with operation $*$ and $H$ with operation $\diamond$ , an isomorphism is a bijection $f:G\to H$ satisfying

f(a*b)=f(a)\diamond f(b)\qquad\text{for every }a,b\in G

A bijection is both one-to-one, so different inputs have different images, and onto, so every target element is an image.

If an isomorphism exists, write $G\cong H$ , read “ $G$ is isomorphic to $H$ ”.

Both requirements matter. Swapping the smaller match to $0\mapsto-1$ , $1\mapsto1$ still gives a bijection, but now

f(0+_2 0)=-1\ne1=f(0)f(0)

so it is not an isomorphism. Conversely, sending both $0,1$ to $1$ preserves the operation but is not one-to-one and misses $-1$ . Operation preservation by itself is also insufficient.

A finite pair of groups must have the same order to be isomorphic, because a bijection pairs up their elements. This is a necessary size check, followed by a structural check. To prove a particular mapping works, establish the law for every pair, by a general argument or a complete Cayley-table comparison. One successful product is only a check.

Key idea

An isomorphism is a reversible relabelling that preserves every product. Equal sizes or a visually similar list of elements do not establish it.

Separate the two requirements in a new example.

Constructing and proving an isomorphism

For cyclic groups of the same finite order, match a generator in the first to a generator in the second, and match their corresponding powers. The cycle lengths make this a bijection; the exponent-addition rule then proves operation preservation. Writing only the generator's image leaves the rest of the mapping and its proof unfinished.

Let $R=\{e,r,r^2,r^3\}$ be square rotations under composition, where $r$ is an anticlockwise quarter-turn and $r^4=e$ . Let $U=(\{1,2,3,4\},\times_5)$ . Our convention is $uv=u\circ v$ , with the right-hand transformation applied first.

Worked example 1: matching the generator to two

Question. Construct an isomorphism $\phi:R\to U$ with $\phi(r)=2$ , and prove it is an isomorphism.

Plan. Match powers of $r$ with powers of $2$ modulo $5$ , then prove the law for arbitrary exponent pairs.

\begin{array}{c|rrrr} x&e&r&r^2&r^3\\\hline \phi(x)&1&2&4&3 \end{array}

These four distinct images are exactly the target elements, so $\phi$ is one-to-one and onto. Its rule is $\phi(r^j)\equiv2^j\pmod5$ , taking the representative in $\{1,2,3,4\}$ . It is consistent when exponents differ by $4$ , since $2^4\equiv1\pmod5$ .

For any $i,j\in\{0,1,2,3\}$ ,

\begin{aligned} \phi(r^i\circ r^j)&=\phi(r^{i+j})\\ &\equiv2^{i+j}\pmod5\\ &\equiv2^i2^j\pmod5\\ &\equiv\phi(r^i)\phi(r^j)\pmod5 \end{aligned}

Thus $\phi(r^i\circ r^j)=\phi(r^i)\times_5\phi(r^j)$ : both sides are the representatives in $\{1,2,3,4\}$ of the same residue, so they are equal as group elements.

Answer. The displayed mapping is a bijection and preserves every product; therefore $R\cong U$ .

Check. Independently check a product that crosses the cycle boundary: $r^2\circ r^3=r$ , whose image is $2$ . The image product is $4\times_5 3=2$ , as required. Also the inverse pair $r,r^3$ maps to the inverse pair $2,3$ .

Keep both groups fixed and change only which target generator receives $r$ . An isomorphism need not be unique.

Worked example 2: matching the generator to three

Question. Construct and prove an isomorphism $\psi:R\to U$ with $\psi(r)=3$ .

Plan. Use powers of $3$ , and verify the same two requirements.

\begin{array}{c|rrrr} x&e&r&r^2&r^3\\\hline \psi(x)&1&3&4&2 \end{array}

Every target element occurs exactly once, so the mapping is a bijection. Since $3^4\equiv1\pmod5$ , the rule $\psi(r^j)\equiv3^j\pmod5$ , taking the allowed representative, respects the rotation's four-step cycle. For all $i,j\in\{0,1,2,3\}$ ,

\psi(r^i\circ r^j) \equiv3^{i+j} \equiv3^i3^j \equiv\psi(r^i)\psi(r^j)\pmod5

Taking the allowed remainder representatives gives
$\psi(r^i\circ r^j)=\psi(r^i)\times_5\psi(r^j)$ .

Answer. $\psi$ is another isomorphism, proving $R\cong U$ .

Check. The square $r\circ r=r^2$ has image $4$ , agreeing with $3\times_5 3=4$ . The image of $r^3=r^{-1}$ is $2$ , and $3\times_5 2=1$ , confirming the inverse correspondence separately.

Matching $r$ to $4$ would fail: powers of $4$ alternate $4,1$ , so two target elements would be missed. The power rule would preserve products, but the map would not be a bijection. This is why the chosen image must be a generator of the target cyclic group.

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DG12 - Isomorphism of finite groups

Bijections that preserve the operation

Constructing and proving an isomorphism

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