Hard Further Maths Groups Questions

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.

Apply Lagrange's theorem

The order of every subgroup must be a divisor of $8$.

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

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

Exhibit a subgroup of order $1$

The closure of a suitable subset really does have order $1$, so this divisor does occur.

Exhibit a subgroup of order $2$

The closure of a suitable subset really does have order $2$, so this divisor does occur.

Exhibit a subgroup of order $4$

The closure of a suitable subset really does have order $4$, so this divisor does occur.

Exhibit a subgroup of order $8$

The closure of a suitable subset really does have order $8$, so this divisor does occur.

Collect the orders that actually occur

This is the complete list of subgroup orders.

Note that Lagrange's theorem is only a restriction

The converse of Lagrange's theorem is false in general, so each order must be checked.

Note the trivial subgroups

These two always appear in the list.

Count the subgroups of each order

There may be several subgroups of the same order.

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 $\left|G\right|=8$, 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 not abelian

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

Select the complete list of subgroup orders

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

Find the order of $G$

Isomorphic groups must have the same order, which already rules out some options.

Work out the order of every element

The list of element orders is preserved by any isomorphism.

Record the order profile

This profile is the fingerprint used to identify the group.

Decide whether $G$ is cyclic

A group is cyclic exactly when some element generates the whole group.

Decide whether $G$ is abelian

$G$ is abelian, which separates it from the remaining options.

Match the profile with one of the standard groups

The two groups have identical order profiles, and an explicit isomorphism can be built.

Reject the non-cyclic options of the same order

The presence or absence of an element of maximal order settles this.

Construct the isomorphism explicitly

Matching a generator (or generating pair) with one of the same order builds the map.

Check the isomorphism preserves the operation

The whole Cayley table is carried across, entry by entry.

List the elements of $G$

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

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.

Note the identity element of $G$

The identity of $G$ is $\begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix}$; 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 cyclic

A single element generates the whole group, so $G\cong C_{4}$.

Recall the four group axioms

A set with a binary operation is a group precisely when all four axioms hold.

Select the correct description

This is the only listed group with the same structure as $G$.

Check that the given image is admissible

An isomorphism preserves orders, so the generator must map to an element of the same order.

Use the homomorphism property on the power

Repeated use of $\phi\left(xy\right)=\phi\left(x\right)\phi\left(y\right)$ turns a power into a power of the image.

Substitute the given image

The image of the generator is the only piece of information needed.

Reduce the index modulo the order of the group

Since $\omega^{8}=1$, the exponent may be reduced modulo $8$.

Tabulate the whole isomorphism as a check

Every element of $H$ appears exactly once, so $\phi$ really is a bijection.

Note that the whole isomorphism is now determined

Once the image of the generator is fixed, every other image is forced.

Check that $\phi$ is a bijection

The image of the generator generates $H$, so $\phi$ is onto and hence bijective.

Check the homomorphism property

The operation is respected for every pair of elements.

Note that the identity maps to the identity

This is forced by the homomorphism property.

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 $\left|G\right|=8$, 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.

Note the identity element of $G$

The identity of $G$ is $1$; 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.

Select the image of the power

This is the element of $H$ that the isomorphism sends $a^{5}$ to.

Compare the orders of the groups in each pair

Isomorphic groups must have the same order, which disposes of some pairs at once.

Work out the order profile of the first group of the correct pair

The number of elements of each order is preserved by any isomorphism.

Work out the order profile of the second group

The two profiles agree exactly, so an isomorphism is possible.

Construct the isomorphism

Sending a generator to a generator and extending to powers gives a bijection that respects the operation.

Refute the pair $\left(\mathbb{Z}_{4},+_{4}\right)$, $\left(U_{8},\times_{8}\right)$

The two order profiles differ, and an isomorphism must preserve the order of every element, so no isomorphism can exist.

Refute the pair $\left(D_{3},\circ\right)$, $\left(\mathbb{Z}_{6},+_{6}\right)$

The two order profiles differ, and an isomorphism must preserve the order of every element, so no isomorphism can exist.

Refute the pair $\left(U_{8},\times_{8}\right)$, $\left(C_{5},\circ\right)$

Groups of different orders can never be isomorphic, because an isomorphism is a bijection.

Refute the pair $\left(D_{4},\circ\right)$, $\left(\mathbb{Z}_{8},+_{8}\right)$

The two order profiles differ, and an isomorphism must preserve the order of every element, so no isomorphism can exist.

Recall the definition of an isomorphism

The map must be one-to-one, onto, and carry the whole operation table across.

Note the cyclic test

Both groups of the correct pair are cyclic of the same order.

Note the abelian test

An abelian group can never be isomorphic to a non-abelian one.

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.

Select the isomorphic pair

Both are cyclic of the same order, so the two groups have identical structure.