1.) Let M be the 2 × 2 matrices with entries in Z/2Z. (So, the entries are 0 or 1, and we go modulo 2 in each entry). ne can compute their determinant, by proceding as usual but reading the answer as “modulo 2”. Let G be those matrices in M that are symmetric, and whose determinant is nonzero (so, it is not 0 mod 2 and therefore must be 1 mod 2, since there are no other elements in Z/2Z). You may assume as given that G is a group with respect to matrix multiplication. Write down all elements (there are fewer than 8) of G explicitly, and find the center of G.

2.) Let H be a subgroup of G and pick x ∈ G. Let H′ = xHx−1
= {xhx−1 |h ∈ H}. Show that H′ is a subgroup of G (show: if
y_{1}H′ then also y −1 ∈ H′ , and show also that if y, z ∈
H′ then y · z ∈ H′ ). Show that the relabeling h ↔ xhx−1 makes the
multiplication tables of H and H′ correspond. We say H, H′ are
conjugate subgroups. Explain why H is conjugate to itself (use x =
e). Show that if G is Abelian then H has only itself as
conjugate.

3.) How many elements does the symmetry group of the cube have?

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Given that M is the set of 2 \times 2 matrices with entries in \mathbb{Z} / 2 \mathbb{Z}, and G is those matrices in in M that are symmetric and whose determinate determinant is norzero.Mcontwing 16 elementr, sinie eain row of a matrix can be chosen trom 4 ordered puir (0,0),(0,1),(1,0),(1,1).Let A \in M be invertible. Since \operatorname{det} A \notin 0, the first row is of 2^{2}-1 non-zero elemenis of \mathbb{Z} / 2 \mathbb{Z} \times \mathbb{Z}. If (a, b)[\neq(0,0)] be the first row of A, cannot be 20(a, b),(0,0).Thus the first row of A can be chosen in \left(2^{2}-1\right) ways corresponding to each such choice, the second row can be chose in \left(2^{2}-2\right) waysHence the total number invertible matrices are\left(2^{2}-1\right)\left(2^{2}-2\right)=6They aresymnetric.G\left\{\begin{array}{l}\text { Gi }\end{array}\left\{\begin{array}{ll}0 & 1 \\1 & 0\end{array}\right),\left(\begin{array}{ll}0 & 1 \\1 & 1\end{array}\right),\left(\begin{array}{ll}1 & 0 \\0 & 1\end{array}\right), \quad\left(\begin{array}{ll}1 & 1 \\1 & 0\end{array}\right)\right\} .=\beta .Given G is group with respect to matrix multiplication.But G cannot be order 4, as \left(\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right)\left(\begin{array}{ll}1 & 1 \\ 1 & 0\end{array}\right)=\left(\begin{array}{ll}1 & 0 \\1 & 1\end{array}\right) \& B .G is \left\{\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right),\left(\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right)\right\} or \left.\left\{\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right),\left(\begin{array}{ll}1 & 1 \\ 1 & 0\end{array}\right)\right\} \quad 0 rcenter of \left.G\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right),\left(\begin{array}{ll}0 & 1 \\ 1 & 1\end{array}\right)\right\}. is G isef.2. Given H \leq G . \& \quad H^{\prime}=x H x^{-1}=\left\{x h x^{-1}: h \in H\right\}. elearly, H^{\prime} is nonemtly as identity e \in H^{\prime}.Let p \in H^{\prime} & q \in H^{\prime}p=x h_{1} x^{-1} \text { \& } q=x h_{2} x^{-1} \text { for some } h_{1}, h_{2} \in H \text {. }\begin{aligned}p q & =\left(x h\left(x^{-1}\right)\left(x h_{2} x^{-1}\right) .\right. \\& =\left(x h_{1}\right)\left(x^{-1} x\right)\left(h_{2} x^{-1}\right) \\& =x\left(h_{1} h_{2}\right) x^{-1} \\& =x h_{3} x^{-1} \text { where } h_{1} h_{2}=h_{3} \in H \text { as } H \leq G . \\p q & H^{\prime} . \\p^{\prime} & =\left(x h_{1} x^{-1}\right)^{-1}=x h_{1}^{-1} x^{-1} \in x H x^{-1} \text { as } H \leq G .\end{aligned}Thus p, q \in x H x^{-1} \Rightarrow p q \in x H_{x}^{-1} \quad \& p \in x H_{x}^{-1} \Rightarrow p^{-1} \in x H_{x}^{-1} It follows that H^{\prime}=x+x^{-1} is a subgroup of G. Let us defire a map \phi: H \rightarrow x H x^{-1} by\phi(h)=x h x tclearly, \phi is well-defined.\begin{aligned}\phi\left(h_{1} h_{2}\right)=x\left(h_{1} h_{2}\right) x^{-1} & =\left(x h_{1} x^{-1}\right)\left(x h_{2} x^{-1}\right) \\& =\phi\left(h_{1}\right) \phi\left(h_{2}\right) \quad \forall h_{1}, h_{2} \in H .\end{aligned}This shows that \phi is a homomorphism.\begin{array}{l}\phi\left(h_{1}\right)=\phi\left(h_{2}\right) \\\Rightarrow x h_{1} x^{-1}=x h_{2} x^{-1} \\\Rightarrow \quad h_{1}=h_{2} \quad[B y \text { cancellation law }]\end{array}Thus \phi is iniective.By the definition of \phi, \phi is onto.Hence \phi is an isomorphism& It follows that relabelling h \leftrightarrow x h x^{-1} makes the anultiplication tables of H and H^{\prime} correspond.H=e H e^{-1}, \mathrm{O}, H \text { is conjugate to itself. }Let G is abelian.\begin{aligned}x+H H_{x}^{-1} & =\left\{x h x^{-1}: h \in H\right\} \\& =\left\{x x^{-1} h: h \in H\right] \quad[A s \text { G is abe lion] } \\& =\{h c h \in H\} \\\therefore \text { This gives } & =H .\end{aligned}This gives the =H proof.3. The group of rigid motions of a cube is S_{4}. So \left|5_{4}\right|=24.The rotationas group of cuble only has 24 elementBut the oymmetric group of the cube is not the wame. an the group of ratations of cuble about some line There are als o reflection There are atso 24 reflechionHence symmetric group of a cube han (24+2.4)=48 dements ...