## (i)

| $\mathbf{X} = \frac{1}{\sqrt{3}}\begin{pmatrix} 2 & -1 \\ 1 & 1 \end{pmatrix} \Rightarrow \mathbf{X}^2 = \begin{pmatrix} 1 & -1 \\ 1 & 0 \end{pmatrix}$, $\mathbf{X}^3 = \frac{1}{\sqrt{3}}\begin{pmatrix} 1 & -2 \\ 2 & -1 \end{pmatrix}$ | M1, A1 | Attempt at powers of X (possibly BC); Correct up to $\mathbf{X}^6$ (NB – not all needed) |
| $\mathbf{X}^4 = \begin{pmatrix} 0 & -1 \\ 1 & -1 \end{pmatrix}$, $\mathbf{X}^5 = \frac{1}{\sqrt{3}}\begin{pmatrix} -1 & -1 \\ 1 & -2 \end{pmatrix}$, $\mathbf{X}^6 = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}$ | A1 | Correct up to $\mathbf{X}^6$; Candidates may calculate each of these BC |
| Note that $\mathbf{X}^7 = -\mathbf{X}$ etc. so that none of $\mathbf{X}^7$ to $\mathbf{X}^{11}$ is equal to $\mathbf{I}$ $\Rightarrow \mathbf{X}^{12} = \mathbf{I}$ and so $n(M) = 12$ because the powers of X now cycle every 12th power but no sooner | A1, E1 [4] | Candidates may calculate each of these BC; Correctly deduced; Clear statement needed to justify $n(M) = 12$ |

## (ii)

| The set is closed (each element is one of the 12 powers of X); matrix multiplication is known to be associative; the multiplicative identity I is in M; and the inverse of $\mathbf{X}^k$ is $\mathbf{X}^{12-k}$ for all suitable $k$ … so we have a group $G$ | B1, B1, B1, B1 | One for each group axiom |
| The group is cyclic since it is generated by **X** (or **X**³ or **X**⁷ or **X**¹¹) **oe** | E1 [5] | Only one generator needs to be mentioned here |

## (iii)

| Proper subgroups of $G$ are $H_2 = [\mathbf{X}^6, \mathbf{I}]$, $H_3 = [\mathbf{X}^4, \mathbf{X}^8, \mathbf{I}]$, $H_4 = [\mathbf{X}^3, \mathbf{X}^6, \mathbf{X}^9, \mathbf{I}]$, $H_6 = [\mathbf{X}^2, \mathbf{X}^4, \mathbf{X}^6, \mathbf{X}^8, \mathbf{X}^{10}, \mathbf{I}]$ | B1, B1, B1 | Any two correct; All four correct (+ no extras – ignore inclusion of {I} and/or $G$) |
| The above subgroups are those generated as follows: $H_2 = (\mathbf{X}^6), H_3 = (\mathbf{X}^4) = (\mathbf{X}^8), H_4 = (\mathbf{X}^3) = (\mathbf{X}^9)$ and $H_6 = (\mathbf{X}^2) = (\mathbf{X}^{10})$ and are thus unique. Mention that *Lagrange's Theorem* states that $\mid H \mid \mid \mid G \mid$ so there are no subgroups of orders 5, 7, 8, 9, 10 or 11 | B1, E1 | Condone omission to mention that any subgroup containing $\mathbf{X}^3$ or $\mathbf{X}^7$ or $\mathbf{X}^{11}$ is $G$ (since they are generators) |
| **OR** For listing the subgroups (as above) The elements have orders [12, 6, 4, 3, 2, 12, 3, 2, 12, 3, 2, 1] and one can argue from the orders of elements that (e.g.): a subgroup of order 2 must contain **I** and an element of order 2 ⇒ $H_2$ uniquely; a subgroup of order 3 must contain **I** and an element of order 3 ⇒ $H_3$ uniquely, since there are only two such elements and they are an inverse-pair; similarly for $H_6$ Then as above for no subgroups of orders 5, 7, 8, 9, 10 or 11 | B1, E1, B1 | |