Further Maths Groups Practice Questions

List the elements of $G$

Writing out the element set is the safest way to avoid missing or repeating an element.

Count the elements

The order of a finite group is the number of elements in its element set.

State the order of $G$

The order of a group is the number of elements it contains.

Apply Lagrange's theorem

The order of every subgroup must be a divisor of $6$.

List the divisors of $\left|G\right|$

These are the only orders that a subgroup could possibly have.

Collect the orders that actually occur

This is the complete list of subgroup orders.

Select the complete list of subgroup orders

These are exactly the orders that occur among the subgroups of $G$.

Check closure

Reducing modulo $7$ always lands back inside the set, so closure holds.

Check associativity

Associativity is inherited from ordinary multiplication of integers, so it holds.

Check for an identity element

The residue $1$ is an identity, so the identity axiom holds.

Check for inverses

No residue multiplies $0$ up to the identity, so $0$ has no inverse.

Identify the axiom that fails

Three of the four axioms hold; only the existence of inverses breaks down.

Note that the operation is also commutative

Commutativity is not a group axiom anyway, and here it holds in any case.

Select the axiom that fails

The set is closed and associative and has an identity, but $0$ has no inverse.

State what the identity must do

The identity must leave every element of the group unchanged.

Test the candidates against one element

Only an element that fixes $2$ can possibly be the identity.

Test the surviving candidate against every element

This element leaves every element of $G$ unchanged, so it is the identity.

Note that the identity is unique

A group cannot have two different identity elements.

Note the row of the Cayley table

In a Cayley table the identity's row and column are copies of the element list.

Note that the identity is the only element of order $1$

Every other element needs at least two applications to return to the identity.

List the elements of $G$

Having the whole element list to hand makes every exhaustive check routine.

Record the order of every element of $G$

Every one of these orders divides $\left|G\right|=4$, exactly as Lagrange's theorem requires.

Note the divisors of $\left|G\right|$

By Lagrange's theorem only these numbers can be orders of subgroups, and only these can be orders of elements.

Note that $G$ is abelian

Every pair of elements commutes, so the Cayley table of $G$ is symmetric about its leading diagonal.

Select the identity element

This element leaves every element of $G$ unchanged under the operation.

Recall the subgroup test

A subgroup must contain the identity and be closed under the operation and inverses.

Check that the correct option contains the identity

A subset without the identity cannot be a subgroup.

Check closure for the correct option

Every product of two elements of the subset is back in the subset.

Check the order against Lagrange's theorem

The order of a subgroup must divide the order of the group.

Reject \left\{e,\ \left(3\,4\right),\ \left(1\,2\right),\ \left(1\,2\,3\right)\right\}

This subset is not closed under the group operation.

Reject \left\{e,\ \left(1\,2\right),\ \left(1\,2\,3\right),\ \left(1\,3\,2\right)\right\}

This subset is not closed under the group operation.

Reject \left\{e,\ \left(1\,2\right),\ \left(1\,3\right),\ \left(1\,4\right)\right\}

This subset is not closed under the group operation.

Note that a subgroup is a group in its own right

Associativity is inherited from $G$, so only closure, identity and inverses need checking.

Record the order profile of $G$

The counts add up to $\left|G\right|=24$, so no element has been missed.

Note the divisors of $\left|G\right|$

By Lagrange's theorem only these numbers can be orders of subgroups, and only these can be orders of elements.

Note that $G$ is not abelian

One failing pair is enough: the order of the factors matters throughout $G$.

Note the identity element of $G$

The identity of $G$ is $e$; every order calculation and every inverse is measured against it.

Note the self-inverse elements of $G$

These are the elements of order $1$ or $2$; they sit on the leading diagonal of the Cayley table as the identity.

Note that $G$ is not cyclic

No single element generates $G$, so $G$ is not isomorphic to $C_{24}$.

Group the permutations of $G$ by cycle type

Two permutations have the same order whenever they have the same cycle type.

Select the subgroup

This is the only option that passes the subgroup test.