\(\{ G,{\text{ }} * \} \) is a group of order \(N\) and \(\{ H,{\text{ }} * \} \) is a proper subgroup of \(\{ G,{\text{ }} * \} \) of order \(n\).

(a)     Define the right coset of \(\{ H,{\text{ }} * \} \) containing the element \(a \in G\).

(b)     Show that each right coset of \(\{ H,{\text{ }} * \} \) contains \(n\) elements.

(c)     Show that the union of the right cosets of \(\{ H,{\text{ }} * \} \) is equal to \(G\).

(d)     Show that any two right cosets of \(\{ H,{\text{ }} * \} \) are either equal or disjoint.

(e)     Give a reason why the above results can be used to prove that \(N\) is a multiple of \(n\).

(a)     the right coset containing \(a\) has the form \(\{ ha|h \in H\} \)     A1

[1 mark]

Note: From here on condone the use of left cosets.

(b)     let \(b\), \(c\) be distinct elements of \(H\). Then, given \(a \in G\), by the Latin square property of the Cayley table, \(ba\) and \(ca\) are distinct     A1

therefore each element of \(H\) corresponds to a unique element in the coset which must therefore contain \(n\) elements     R1

[2 marks]

(c)     let \(d\) be any element of \(G\). Then since \(H\) contains the identity \(e\), \(ed = d\) will be in a coset     R1

therefore every element of \(G\) will be contained in a coset which proves that the union of all the cosets is \(G\)     R1

[2 marks]

(d)     let the cosets of \(b\) and \(c{\text{ }}(b,{\text{ }}c \in G)\) contain a common element so that

\(pb = qc\) where \(p,{\text{ }}q \in H\). Let \(r\) denote any other element \( \in H\)     M1

then

\(rb = r{p^{ - 1}}qc\)     A1

since \(r{p^{ - 1}}q \in H\), this shows that all the other elements are common and the cosets are equal     R1

since not all cosets can be equal, there must be other cosets which are disjoint     R1

[4 marks]

(e)     the above results show that \(G\) is partitioned into a number of disjoint subsets containing \(n\) elements so that \(N\) must be a multiple of \(n\)     R1

[1 mark]