Xa7, Xa8 - Isomorphic groups | OCR A-Level Further Mathematics B (MEI)

Groups can describe different objects while obeying exactly the same combination rules. Xa7 recognises that common structure; Xa8 specifies an isomorphism, proves it preserves the operation, and uses element orders to rule out impossible correspondences.

Recognise the same operation behind different labels

What could adding one on a four-position dial have in common with multiplying by $i$ or rotating a square through a quarter-turn? Each action takes four repetitions to return to the identity. More is true: we can pair up the elements so that every combination agrees.

You already know

A finite cyclic group is generated by one element, whose order equals the group order. Subgroups use the original operation. A composition $S\circ T$ performs $T$ first, then $S$ , matching matrix multiplication $\mathbf M_S\mathbf M_T$ .

Pair the additive residues $0,1,2,3$ with $1,i,-1,-i$ , respectively. For example, $3+_4 2=1$ corresponds to $(-i)(-1)=i$ . The identity changes its label from zero to one, but its role is preserved.

Isomorphism

An isomorphism from $(G,*)$ to $(H,\circ)$ is a bijection $\phi:G\to H$ that preserves the operation:

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

A bijection assigns each element of $G$ to exactly one element of $H$ , with every element of $H$ used exactly once. If such a map exists, the groups are isomorphic.

There are two jobs: match the elements one-to-one, and show that combining before mapping agrees with mapping before combining. A list pairing equal-sized sets deals with the first job only. One successful product is a check, not a proof for all products.

[DIAGRAM: asset_name: One four-step structure in three models; asset_slug: h645_b142_correspondence; caption: Identities and successive generator steps correspond across residues, roots of unity and square rotations; description: Three rows list additive residues zero, one, two, three; complex numbers one, i, minus one, minus i; and square rotations I, R, R squared, R cubed. Vertically aligned arrows pair each column. A note shows three plus two modulo four equals one corresponding to minus i times minus one equals i and R cubed composed with R squared equals R.]
Identities and successive generator steps correspond across residues, roots of unity and square rotations

Isomorphisms preserve identities, inverses and element orders. For example, $\phi(e_G)\circ\phi(a)=\phi(e_G*a)=\phi(a)$ ; surjectivity lets $\phi(a)$ run through all of $H$ , so $\phi(e_G)=e_H$ . Similarly, $\phi(a^{-1})=\phi(a)^{-1}$ . A first identity return must occur at the same step in both groups, because the map is one-to-one.

These definitions and properties are not printed in the formula booklet. They explain why the label $0$ can legitimately correspond to $1$ : preserve the identity's role rather than its written value.

Specify an isomorphism by matching generators

For finite cyclic groups, one generator determines the whole correspondence. If $G=\langle g\rangle$ and $H=\langle h\rangle$ both have order $n$ , send $g^r$ to $h^r$ . The exponents are read modulo $n$ , so the rule is well-defined: two expressions for the same source element differ by a multiple of $n$ and give the same target element.

Finite cyclic groups of equal order

Two finite cyclic groups of the same order $n$ are isomorphic. Given generators $g,h$ , an isomorphism is

\phi(g^r)=h^r,\qquad r=0,1,\ldots,n-1

It is a bijection because these are all the distinct elements of each group. It preserves products because

\phi(g^r g^s)=h^{r+s}=h^r h^s=\phi(g^r)\phi(g^s)

Not in the formula booklet: learn it.

In an additive source group, a generator's repeated sums take the place of powers. Its identity must map to the target identity. Choose a generator of the target, not an element that produces only a proper subgroup.

Worked example 1: addition becomes modular multiplication

Question. Let $G=(\{0,1,\ldots,5\},+_6)$ and let $H=(\{1,2,3,4,5,6\},\times_7)$ , where $\times_7$ is multiplication modulo $7$ . Specify an isomorphism sending $1\in G$ to $3\in H$ , and justify it.

Plan. List the powers of $3$ modulo $7$ , then use exponent addition to prove the product condition.

$k\in G$	$0$	$1$	$2$	$3$	$4$	$5$
$\phi(k)=3^k\pmod7$	$1$	$3$	$2$	$6$	$4$	$5$

\begin{aligned} 3^6&\equiv1\pmod7 &&\text{first return; six distinct powers}\\ \phi(a+_6 b)&\equiv3^{a+b}\pmod7 &&\text{reducing exponent by }6\text{ changes nothing}\\ &\equiv3^a3^b\equiv\phi(a)\times_7\phi(b)\pmod7 &&\text{operation preserved} \end{aligned}

Why this step? The first row uses addition modulo six, while the second uses multiplication modulo seven; ordinary equality of numerical labels is irrelevant.

Answer. The table, or $\phi(k)$ equal to the remainder of $3^k$ modulo $7$ , specifies an isomorphism. It is bijective and preserves every product as shown.

Check. $4+_6 5=3$ maps to $6$ ; $\phi(4)\times_7\phi(5)=4\times_7 5=6$ agrees.

Change only the image of the source generator: send $1$ to $5$ instead of $3$ . A different generator can give a different isomorphism.

Worked example 2: a second valid correspondence

Question. Construct and justify this second isomorphism $\psi:G\to H$ .

Plan. Repeat the generator construction with powers of $5$ , checking that all target elements appear.

$k\in G$	$0$	$1$	$2$	$3$	$4$	$5$
$\psi(k)=5^k\pmod7$	$1$	$5$	$4$	$6$	$2$	$3$

\begin{aligned} 5^6&\equiv1\pmod7 &&\text{six distinct powers before return}\\ \psi(a+_6 b)&\equiv5^{a+b}\equiv5^a5^b\pmod7 &&\text{same general exponent argument}\\ &=\psi(a)\times_7\psi(b) &&\text{equality of the chosen residues} \end{aligned}

Why this step? A changed generator image forces new images for every element; changing only one table entry would generally destroy operation preservation.

Answer. $\psi(k)$ is the remainder of $5^k$ modulo $7$ . The table gives a bijection, and the displayed general identity proves it is an isomorphism.

Check. It differs from $\phi$ , since $\psi(1)=5$ whereas $\phi(1)=3$ , but both send zero to one.

If we tried $1\mapsto2$ , powers $2,4,1$ would repeat after three steps, missing half of $H$ . The proposed map could not be bijective. Matching the source generator's order prevents that error.

Watch out

Wrong: $\phi(0)=0$ for the map from addition modulo six to multiplication modulo seven.

Why: an isomorphism preserves the identity's role, not its numerical label; the target identity is $1$ .

Right: $\phi(0)=1$ . Choose the image of a generator, then extend the map consistently to every element.

Use the same identity and generator logic for a target written with complex roots instead of modular residues.

Match transformations to matrices

A linear transformation's matrix records what it does to every vector. Therefore a faithful assignment of distinct transformations to their matrices is more than a resemblance between tables: composition corresponds to multiplication for every input vector.

For square rotations let $R$ be an anticlockwise quarter-turn, and put

\mathbf J=\begin{pmatrix}0&-1\\1&0\end{pmatrix},\qquad \mathbf J^2=-\mathbf I,\quad\mathbf J^3=-\mathbf J,\quad\mathbf J^4=\mathbf I

The two groups are $G=\{I,R,R^2,R^3\}$ under composition and $H=\{\mathbf I,\mathbf J,-\mathbf I,-\mathbf J\}$ under multiplication. Their identity symbols denote different kinds of objects.

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Xa7, Xa8 - Isomorphic groups

Recognise the same operation behind different labels

Specify an isomorphism by matching generators

Match transformations to matrices

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