State Lagrange’s theorem.

Verify that the inverse of \(a * {b^{ - 1}}\) is equal to \(b * {a^{ - 1}}\).

Let \(\{ H,{\rm{ }} * {\rm{\} }}\) be a subgroup of \(\{ G,{\rm{ }} * {\rm{\} }}\). Let \(R\) be a relation defined on \(G\) by

\[aRb \Leftrightarrow a * {b^{ - 1}} \in H.\]

Prove that \(R\) is an equivalence relation, indicating clearly whenever you are using one of the four properties required of a group.

Let \(\{ H,{\rm{ }} * {\rm{\} }}\) be a subgroup of \(\{ G,{\rm{ }} * {\rm{\} }}\) .Let \(R\) be a relation defined on \(G\) by

\[aRb \Leftrightarrow a * {b^{ - 1}} \in H.\]

Show that \(aRb \Leftrightarrow a \in Hb\), where \(Hb\) is the right coset of \(H\) containing \(b\).

Let \(\{ H,{\rm{ }} * {\rm{\} }}\) be a subgroup of \(\{ G,{\rm{ }} * {\rm{\} }}\) .Let \(R\) be a relation defined on \(G\) by

\[aRb \Leftrightarrow a * {b^{ - 1}} \in H.\]

It is given that the number of elements in any right coset of \(H\) is equal to the order of \(H\).

Explain how this fact together with parts (c) and (d) prove Lagrange’s theorem.

in a finite group the order of any subgroup (exactly) divides the order of the group     A1A1

[2 marks]

METHOD 1

\((a * {b^{ - 1}}) * (b * {a^{ - 1}}) = a * {b^{ - 1}} * b * {a^{ - 1}} = a * e * {a^{ - 1}} = a * {a^{ - 1}} = e\)     M1A1A1

Note:     M1 for multiplying, A1 for at least one of the next 3 expressions,

A1 for \(e\).

Allow \((b * {a^{ - 1}}) * (a * {b^{ - 1}}) = b * {a^{ - 1}} * a * {b^{ - 1}} = b * e * {b^{ - 1}} = b * {b^{ - 1}} = e\).

METHOD 2

\({(a * {b^{ - 1}})^{ - 1}} = {({b^{ - 1}})^{ - 1}} * {a^{ - 1}}\)     M1A1

\( = b * {a^{ - 1}}\)A1

[3 marks]

\(a * {a^{ - 1}} = e \in H\;\;\;\)(as \(H\) is a subgroup)     M1

so \(aRa\) and hence \(R\) is reflexive

\(aRb \Leftrightarrow a * {b^{ - 1}} \in H\). \(H\) is a subgroup so every element has an inverse in \(H\) so

\({(a * {b^{ - 1}})^{ - 1}} \in H\)     R1

\( \Leftrightarrow b * {a^{ - 1}} \in H \Leftrightarrow bRa\)     M1

so \(R\) is symmetric

\(aRb,{\text{ }}bRc \Leftrightarrow a * {b^{ - 1}} \in H,{\text{ }}b * {c^{ - 1}} \in H\)     M1

as \(H\) is closed \((a * {b^{ - 1}}) * {\text{(}}b * {c^{ - 1}}) \in H\)     R1

and using associativity     R1

\((a * {b^{ - 1}}) * {\text{(}}b * {c^{ - 1}}) = a * ({b^{ - 1}} * b) * {c^{ - 1}} = a * {c^{ - 1}} \in H \Leftrightarrow aRc\)     A1

therefore \(R\) is transitive

\(R\) is reflexive, symmetric and transitive

Note:     Can be said separately at the end of each part.

hence it is an equivalence relation     AG

[8 marks]

\(aRb \Leftrightarrow a * {b^{ - 1}} \in H \Leftrightarrow a * {b^{ - 1}} = h \in H\)     A1

\( \Leftrightarrow a = h * b \Leftrightarrow a \in Hb\)     M1R1

[3 marks]

(d) implies that the right cosets of \(H\) are equal to the equivalence classes of the relation in (c)     R1

hence the cosets partition \(G\)     R1

all the cosets are of the same size as the subgroup \(H\) so the order of \(G\) must be a multiple of \(\left| H \right|\)     R1

[3 marks]

Total [19 marks]

Many students obtained just half marks in (a) for not stating the requirement of the order to be finite.

Part (b) should have been more straightforward than many found.

In part (c) it was evident that most candidates knew what to do, but being a more difficult question fell down on a lack of rigour. Nonetheless, many candidates obtained full or partial marks on this question part.

Part (d) enabled many candidates to obtain, at least partial marks, but there were few students with the insight to be able to answer part (e) satisfactorily.

Part (d) enabled many candidates to obtain, at least partial marks, but there were few students with the insight to be able to answer part (e) satisfactorily.