Date | November 2010 | Marks available | 8 | Reference code | 10N.3srg.hl.TZ0.5 |
Level | HL only | Paper | Paper 3 Sets, relations and groups | Time zone | TZ0 |
Command term | Show that | Question number | 5 | Adapted from | N/A |
Question
Let \(\{ G,{\text{ }} * \} \) be a finite group that contains an element a (that is not the identity element) and \(H = \{ {a^n}|n \in {\mathbb{Z}^ + }\} \), where \({a^2} = a * a,{\text{ }}{a^3} = a * a * a\) etc.
Show that \(\{ H,{\text{ }} * \} \) is a subgroup of \(\{ G,{\text{ }} * \} \).
Markscheme
since G is closed, H will be a subset of G
closure: \(p,{\text{ }}q \in H \Rightarrow p = {a^r},{\text{ }}q = {a^s},{\text{ }}r,{\text{ }}s \in {\mathbb{Z}^ + }\) A1
\(p * q = {a^r} * {a^s} = {a^{r + s}}\) A1
\(r + s \in {\mathbb{Z}^ + } \Rightarrow p * q \in H\) hence H is closed R1
associativity follows since \( * \) is associative on G (R1)
EITHER
identity: let the order of a in G be \(m \in {\mathbb{Z}^ + },{\text{ }}m \geqslant 2\) M1
then \({a^m} = e \in H\) R1
inverses: \({a^{m - 1}} * a = e \Rightarrow {a^{m - 1}}\) is the inverse of a A1
\({({a^{m - 1}})^n} * {a^n} = e\), showing that \({a^n}\) has an inverse in H R1
hence H is a subgroup of G AG
OR
since \((G,{\text{ }} * )\) is a finite group, and H is a non-empty closed subset of G, then \((H,{\text{ }} * )\) is
a subgroup of \((G,{\text{ }} * )\) R4
Note: To receive the R4, the candidate must explicitly state the theorem, i.e. the three given conditions, and conclusion.
[8 marks]
Examiners report
This question was generally answered very poorly, if attempted at all. Candidates failed to realize that the property of closure needed to be properly proved. Others used negative indices when the question specifically states that the indices are positive integers.