8.03l, 8.03m - Isomorphism and abstract groups | OCR A-Level Further Mathematics A

Two groups can have different-looking elements but identical multiplication structure. You will establish or rule out an isomorphism, and use group laws to solve equations, prove identities and complete abstract Cayley tables. The reasoning will depend on the stated algebraic properties rather than on what the element labels happen to resemble.

Showing that groups are isomorphic

Can addition on a four-hour clock really be the same structure as multiplication of complex numbers? Adding one step four times returns to zero; multiplying by $i$ four times returns to $1$ . Matching those cycles is the start of an exact correspondence, not just a visual analogy.

You already know

A finite cyclic group consists of the powers of a generator. An order- $n$ generator has powers $e,a,\ldots,a^{n-1}$ , with exponents reduced modulo $n$ . Isomorphic-looking patterns must still respect the operation, and a group's size alone does not determine its structure.

Label the clock positions $0,1,2,3$ under addition modulo $4$ , and match them to $1,i,-1,-i$ under multiplication. The rule $k\mapsto i^k$ preserves the calculation: adding clock positions corresponds to multiplying their complex images, because $i^{j+k}=i^j i^k$ .

[DIAGRAM: asset_name: A correspondence between two cyclic groups; asset_slug: h245_8_03_isomorphism; caption: Adding one modulo four and multiplying by i trace matching cycles, and the correspondence also preserves every combined product; description: A left panel shows the cycle from 0 to 1 to 2 to 3 and back to 0 under addition modulo 4. A right panel shows the matching cycle from 1 to i to minus 1 to minus i and back to 1 under complex multiplication by i. Matching positions have the same relative placement. A central mapping sends k to i to the power k, with 0 mapped to 1 and 2 mapped to minus 1.]
Adding one modulo four and multiplying by i trace matching cycles, and the correspondence also preserves every combined product

The identities must match: clock $0$ corresponds to multiplicative $1$ . Their different names do not matter.

Isomorphism

Groups $(G,\circ)$ and $(H,*)$ are isomorphic, written $G\cong H$ , if there is a one-to-one correspondence $f$ from all elements of $G$ onto all elements of $H$ that preserves the operation:

f(a\circ b)=f(a)*f(b)\qquad(a,b\in G)

The correspondence is an isomorphism. Equivalently, consistently rename the row headings, column headings and entries of one Cayley table to obtain the other table. Equal group orders alone do not establish an isomorphism.

For finite cyclic groups of the same order, match one generator to another and extend by powers. This gives a correspondence because the powers in each group are distinct until the same first return; it preserves products because both add exponents modulo that common order. Thus two cyclic groups of the same finite order are isomorphic. The definition and this result are not in the formula booklet.

Worked example 1: matching generators

Question. Let $G=\langle a\rangle$ have order $4$ , and let $H=\{1,i,-1,-i\}$ under multiplication. Show that the correspondence $a^k\mapsto i^k$ is an isomorphism.

Plan. List all images to establish a one-to-one correspondence, then check products using the exponent rule.

\begin{aligned} f(e)&=1,\quad f(a)=i,\quad f(a^2)=-1,\quad f(a^3)=-i &&\text{four distinct images cover }H\\ f(a^r a^s)&=f(a^{r+s})=i^{r+s} &&\text{reduce exponents modulo }4\\ f(a^r)f(a^s)&=i^r i^s=i^{r+s} &&\text{the same result} \end{aligned}

Why check arbitrary powers? They represent every pair of elements in $G$ , so the equality proves product preservation for the whole group.

Answer. The correspondence is bijective and preserves products, so $G\cong H$ .

Check. $a\cdot a^3=e$ maps to $i\cdot(-i)=1$ , the correct identity image.

Change just the image of the generator from $i$ to the other order- $4$ element, $-i$ . The correspondence changes, but the structures still match.

Worked example 2: another isomorphism between the same groups

Question. For the same groups, show that $a^k\mapsto(-i)^k$ is also an isomorphism.

Plan. Recalculate the images and use the same power law for products.

\begin{aligned} g(e)&=1,\quad g(a)=-i,\quad g(a^2)=-1,\quad g(a^3)=i &&\text{again four distinct images}\\ g(a^r a^s)&=(-i)^{r+s} &&\text{the image of the product}\\ g(a^r)g(a^s)&=(-i)^r(-i)^s=(-i)^{r+s} &&\text{the product of the images} \end{aligned}

Why must the generator image have order $4$ ? It must provide four different powers, or the correspondence cannot be one-to-one.

Answer. This is another isomorphism from $G$ to $H$ .

Check. Trying $a\mapsto-1$ instead would give $e\mapsto1$ and $a^2\mapsto1$ , so that attempted correspondence would fail to be one-to-one.

When identifying a group as cyclic, give a generator or establish an element of full group order. Merely describing two groups as cyclic without showing the unfamiliar one has that property leaves the key step unsupported.

Watch out

Wrong: “Both groups have four elements, so they are isomorphic.” Right: preserve the products, for example by showing both are cyclic of order $4$ and matching generators. The non-cyclic Klein four-group also has four elements.

The same correspondence works for another modular multiplication group with a different cycle length.

Ruling out an isomorphism

A mismatch in a structural property rules out every possible renaming at once.

An isomorphism sends the identity to the identity and powers to the corresponding powers: $f(a^k)=f(a)^k$ . Hence it preserves element orders. To justify the last claim, a first return $a^m=e$ gives $f(a)^m=e_H$ ; a smaller return in the image would, by one-to-one correspondence, give a smaller return for $a$ . It also preserves commutativity and subgroup structure. These consequences are not in the formula booklet.

To prove non-isomorphism, compare group orders first, then an informative invariant: a possible element order, the number of elements of some order, cyclicity, or commutativity. Matching one such property does not by itself prove isomorphism.

Worked example 3: an absent element order

Question. Let $A$ be the group of pairs $(x,y)$ , with $x\in\{0,1,2,3\}$ , $y\in\{0,1\}$ , under coordinatewise addition modulo $4$ and $2$ , respectively. Show that $A$ is not isomorphic to $C_8$ .

Plan. Compare the largest possible order of an element with the order of a generator in $C_8$ .

\begin{aligned} |A|&=4\times2=8 &&\text{same size as }C_8\\ 4(x,y)&=(0,0) &&4x\equiv0\pmod4,\quad4y\equiv0\pmod2\\ o((x,y))&\mid4 &&\text{every element returns within four additions} \end{aligned}

But $C_8$ has a generator of order $8$ , while $A$ has no element of that order.

Why compare orders rather than names? An isomorphism may rename every element, but cannot change its first return to the identity.

Answer. $A\not\cong C_8$ , since the existence of an element of order $8$ differs.

Check. $(1,0)$ has order $4$ , so $A$ does have non-trivial cycles; they are simply too short to generate all eight elements.

Keep $A$ , but change just the comparison group to the eight triples of binary coordinates. Now neither group has an element of order $8$ , so use a finer comparison.

Worked example 4: counting elements of order two

Question. Let $B$ be the group of triples $(u,v,w)$ , with each coordinate $0$ or $1$ , under coordinatewise addition modulo $2$ . Show that $A$ , defined above, is not isomorphic to $B$ .

Plan. Count the non-identity elements that return after two additions in each group.

\begin{aligned} 2(x,y)=(0,0)&\ \Longleftrightarrow\ x\in\{0,2\},\ y\in\{0,1\} &&\text{four choices in }A\\ \#\{a\in A:o(a)=2\}&=4-1=3 &&\text{exclude the identity}\\ 2(u,v,w)&=(0,0,0) &&\text{every binary triple returns}\\ \#\{b\in B:o(b)=2\}&=8-1=7 &&\text{exclude the identity} \end{aligned}

Why count non-identity elements separately? An identity that also squares to itself has order $1$ , not $2$ .

Answer. $A\not\cong B$ , because they have different numbers of elements of order $2$ .

Check. The three order- $2$ elements of $A$ are $(0,1),(2,0),(2,1)$ . Alternatively, $A$ contains an element of order $4$ , whereas $B$ does not.

These groups are all abelian and all have eight elements. Those two matches are insufficient; one structural mismatch settles the question.

Watch out

Wrong: “Both groups have one identity and are abelian, so they are isomorphic.” Right: matching selected properties is necessary evidence, not a correspondence. A mismatch proves non-isomorphism; a match alone need not prove isomorphism.

The two familiar order- $6$ structures give a compact further comparison.

Algebra in an abstract group

Group algebra permits cancellation, but a factor must be removed from the side where it occurs.

The identity and the inverse of an element are unique. If $e,f$ were both identities, then $ef=f$ using $e$ , while $ef=e$ using $f$ , so $e=f$ . If $b,c$ were both inverses of $a$ , then

b=be=b(ac)=(ba)c=ec=c

using associativity. These proofs use no commutativity.

Similarly, $ax=ay$ can be cancelled by multiplying on the left by $a^{-1}$ , while $xa=ya$ is cancelled on the right. One must not move a factor past another without a reason.

The inverse of a product

In any group,

(ab)^{-1}=b^{-1}a^{-1}

The factor order reverses. More generally,

(abc)^{-1}=c^{-1}b^{-1}a^{-1}

Not in the formula booklet: learn it.

Continue this lesson.

You’re previewing NovaLearn without an account. Create a free account to read the rest.

Create free account

Last Accessed

Pure Core

Statistics

Mechanics

Discrete Mathematics

Additional Pure Mathematics

8.03l, 8.03m - Isomorphism and abstract groups

Showing that groups are isomorphic

Ruling out an isomorphism

Algebra in an abstract group

Continue this lesson.

Learn faster with Nova