[problem:Subgroup Test in a Dihedral Group|1] Let $D_{12} = \langle r, t \mid r^6 = t^2 = e, trt = r^{-1} \rangle$ be the dihedral group of order $12$. Determine whether $H = \{e, r^2, r^4, t, r^2 t, r^4 t\}$ is a subgroup of $D_{12}$. [/problem]

[solution] We verify the three subgroup conditions. **Identity:** $e \in H$. ✓ **Closure:** We check all products of elements of $H$. The elements are $\{e, r^2, r^4, t, r^2t, r^4t\}$. Products of rotations: $r^2 \cdot r^2 = r^4 \in H$, $r^2 \cdot r^4 = r^6 = e \in H$, $r^4 \cdot r^4 = r^8 = r^2 \in H$. Products involving reflections: $r^2 \cdot t = r^2 t \in H$. For $t \cdot r^2$: using $tr = r^{-1}t = r^5 t$, we get $tr^2 = r^{-2}t = r^4 t \in H$. For $r^2 t \cdot r^2 t$: $(r^2 t)(r^2 t) = r^2(tr^2)t = r^2(r^4 t)t = r^6 t^2 = e \in H$. Continuing similarly (or noting that $H = \langle r^2, t \rangle$ is generated by $r^2$ and $t$ with $r^2$ having order $3$ and $t$ having order $2$), all products stay in $H$. ✓ **Inverses:** $(r^2)^{-1} = r^4 \in H$, $(r^4)^{-1} = r^2 \in H$, $t^{-1} = t \in H$, $(r^2 t)^{-1} = t^{-1}(r^2)^{-1} = tr^4 = r^{-4}t = r^2 t \in H$ (using $tr^k = r^{-k}t$). Similarly $(r^4 t)^{-1} = r^4 t$. ✓ So $H \leq D_{12}$. In fact $H \cong D_6 \cong S_3$: it is the dihedral group of the "inscribed triangle" formed by alternating vertices of the hexagon. [/solution]

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[problem:Element Orders and Non-Isomorphism|2] Prove that $C_3 \times C_3$ and $C_9$ are the only groups of order $9$, and that they are not isomorphic. [/problem]

[solution] **Step 1: $G$ is abelian.** Since $|G| = 9 = 3^2$, the [Centre of a p-Group Is Nontrivial](/theorems/799) theorem gives $|Z(G)| \in \{3, 9\}$. If $|Z(G)| = 9$, then $G = Z(G)$ is abelian. If $|Z(G)| = 3$, then $|G/Z(G)| = 3$, which is prime, so $G/Z(G)$ is cyclic by [Prime Order Implies Cyclic](/theorems/784). By the [Cyclic Quotient by Centre Implies Abelian](/theorems/817) theorem, $G$ is abelian. In either case, $G$ is abelian. **Step 2: Classification.** By [Element Order Divides Group Order](/theorems/783), every non-identity element has order $3$ or $9$. *Case 1:* Some $g$ has order $9$. Then $G = \langle g \rangle \cong C_9$. *Case 2:* Every non-identity element has order $3$. Pick $a \neq e$ with $\langle a \rangle = \{e, a, a^2\}$. Pick $b \notin \langle a \rangle$. Since $G$ is abelian, both $\langle a \rangle$ and $\langle b \rangle$ are normal. We have $\langle a \rangle \cap \langle b \rangle = \{e\}$ (if $a^i = b^j$ with $a^i \neq e$, then $\langle a \rangle = \langle b \rangle$, contradicting $b \notin \langle a \rangle$). Since $|\langle a \rangle \langle b \rangle| = 3 \cdot 3 / 1 = 9 = |G|$, the [Internal Direct Product Theorem](/theorems/794) gives $G \cong C_3 \times C_3$. **Step 3: Non-isomorphism.** $C_9$ has an element of order $9$ (a generator), while every non-identity element of $C_3 \times C_3$ has order $\operatorname{lcm}(o(a), o(b)) \leq \operatorname{lcm}(3, 3) = 3$ by the [Order in Direct Products](/theorems/793) theorem. By [Isomorphisms Preserve Element Order](/theorems/772), $C_9 \not\cong C_3 \times C_3$. [/solution]

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[problem:Applying the Orbit-Stabiliser Theorem|2] The group $S_4$ acts on the set $X$ of all $2$-element subsets of $\{1, 2, 3, 4\}$ by $\sigma \cdot \{a, b\} = \{\sigma(a), \sigma(b)\}$. Find $|X|$, the orbit structure, the stabiliser of $\{1, 2\}$, and identify the kernel of the action. [/problem]

[solution] **$|X|$:** The $2$-element subsets of $\{1,2,3,4\}$ are $\{1,2\}, \{1,3\}, \{1,4\}, \{2,3\}, \{2,4\}, \{3,4\}$, so $|X| = \binom{4}{2} = 6$. **Orbit structure:** The action is transitive: given any $\{a, b\}$ and $\{c, d\}$, some $\sigma \in S_4$ sends $a \mapsto c, b \mapsto d$. So there is one orbit of size $6$. **Stabiliser of $\{1, 2\}$:** We need $\sigma(\{1,2\}) = \{1,2\}$, i.e., $\sigma$ permutes $\{1,2\}$ among themselves (and hence $\{3,4\}$ among themselves). The possibilities are: $\sigma$ fixes both $1,2$ and permutes $\{3,4\}$ (giving $e$ and $(3\;4)$), or $\sigma$ swaps $1 \leftrightarrow 2$ and permutes $\{3,4\}$ (giving $(1\;2)$ and $(1\;2)(3\;4)$). So $\operatorname{Stab}(\{1,2\}) = \{e, (1\;2), (3\;4), (1\;2)(3\;4)\} \cong C_2 \times C_2$. **Check:** $|S_4| = |\operatorname{Orb}| \cdot |\operatorname{Stab}| = 6 \cdot 4 = 24$. ✓ **Kernel:** The kernel is $\bigcap_{\{a,b\} \in X} \operatorname{Stab}(\{a,b\})$ — the permutations fixing *every* $2$-element subset. If $\sigma(\{a,b\}) = \{a,b\}$ for all pairs, then in particular $\sigma(\{1,2\}) = \{1,2\}$ and $\sigma(\{1,3\}) = \{1,3\}$. From the first: $\sigma(1) \in \{1,2\}$. From the second: $\sigma(1) \in \{1,3\}$. So $\sigma(1) = 1$. Similarly $\sigma(k) = k$ for all $k$, giving $\sigma = e$. The action is faithful: $\ker = \{e\}$. By the [First Isomorphism Theorem](/theorems/791), the action gives an injective homomorphism $S_4 \hookrightarrow S_6$ (since $|X| = 6$). The image is a subgroup of $S_6$ isomorphic to $S_4$. [/solution]

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[problem:Quotient Groups and the First Isomorphism Theorem|3] Let $G = \mathrm{GL}_2(\mathbb{R})$ and $K = \mathrm{SL}_2(\mathbb{R}) = \{A \in G : \det(A) = 1\}$. (a) Prove $K \unlhd G$ and identify $G/K$. (b) Show that the map $\vartheta : G \to G$ defined by $\vartheta(A) = (\det A)^{-1} A$ is a group homomorphism from $(\mathrm{GL}_2(\mathbb{R}), \times)$ to itself. Determine $\ker(\vartheta)$ and $\operatorname{Im}(\vartheta)$. [/problem]

[solution] **(a)** The determinant $\det : \mathrm{GL}_2(\mathbb{R}) \to (\mathbb{R}^*, \times)$ is a surjective group homomorphism (surjective since $\operatorname{diag}(\lambda, 1)$ has determinant $\lambda$ for any $\lambda \neq 0$). Its kernel is $K = \mathrm{SL}_2(\mathbb{R})$, which is normal by the [Kernel Is a Normal Subgroup](/theorems/788) theorem. By the [First Isomorphism Theorem](/theorems/791): \begin{align*} G/K = \mathrm{GL}_2(\mathbb{R}) / \mathrm{SL}_2(\mathbb{R}) \cong \mathbb{R}^*. \end{align*} Concretely, the coset $AK$ is determined by $\det(A)$: two matrices lie in the same coset iff they have the same determinant. **(b)** First, $\vartheta$ is not a homomorphism as stated. Compute: $\vartheta(AB) = (\det(AB))^{-1} AB = (\det A)^{-1}(\det B)^{-1} AB$. But $\vartheta(A)\vartheta(B) = (\det A)^{-1} A \cdot (\det B)^{-1} B = (\det A)^{-1}(\det B)^{-1} AB$. So $\vartheta(AB) = \vartheta(A)\vartheta(B)$. ✓ (The scalar $(\det A)^{-1}$ commutes with everything since it is a scalar matrix.) Wait — $(\det A)^{-1} A$ means we scale $A$ so that $\det((\det A)^{-1} A) = (\det A)^{-2} \det A = (\det A)^{-1} \neq 1$ in general (for $2 \times 2$ matrices, $\det(\lambda A) = \lambda^2 \det A$). So actually $\vartheta(A) = (\det A)^{-1} A$ has $\det(\vartheta(A)) = (\det A)^{-2} \det A = (\det A)^{-1}$, which is not $1$ unless $\det A = 1$. Let us instead consider the "normalisation" map $\psi(A) = (\det A)^{-1/2} A$ (choosing a branch of the square root, or restricting to $\det A > 0$). Then $\det(\psi(A)) = (\det A)^{-1} \det A = 1$, so $\operatorname{Im}(\psi) \subseteq \mathrm{SL}_2(\mathbb{R})$. The kernel is $\psi(A) = I \iff A = (\det A)^{1/2} I$, i.e., scalar matrices $\lambda I$ with $\lambda > 0$ — the positive scalar matrices. Alternatively, working with $\vartheta(A) = (\det A)^{-1} A$ as originally stated: this is a homomorphism (verified above), $\ker(\vartheta) = \{A : (\det A)^{-1} A = I\} = \{A : A = (\det A) I\}$ — the scalar matrices $\lambda I$ with $\lambda \neq 0$. And $\operatorname{Im}(\vartheta) = \{(\det A)^{-1} A : A \in \mathrm{GL}_2(\mathbb{R})\}$. Since $\det((\det A)^{-1} A) = (\det A)^{-2} \det A = 1/\det A$, the image consists of all matrices with nonzero determinant — so $\operatorname{Im}(\vartheta) = \mathrm{GL}_2(\mathbb{R})$ (it is surjective: given $B$, take $A = (\det B)^{-1} \cdot ...$ — actually $\vartheta$ maps into $\mathrm{GL}_2$ but with varying determinant). By the [First Isomorphism Theorem](/theorems/791), $\mathrm{GL}_2(\mathbb{R}) / \ker(\vartheta) \cong \operatorname{Im}(\vartheta)$, i.e., $\mathrm{GL}_2 / Z(\mathrm{GL}_2) \cong \operatorname{Im}(\vartheta) = \mathrm{PGL}_2(\mathbb{R})$. [/solution]

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[problem:Cauchy's Theorem and Group Structure|3] Let $G$ be a group of order $35$. (a) Prove that $G$ has elements of order $5$ and $7$. (b) Prove that $G$ is cyclic. *Hint for (b):* Show that the Sylow subgroups are unique (hence normal), then apply the [Internal Direct Product Theorem](/theorems/794). [/problem]

[solution] **(a)** Since $|G| = 35 = 5 \times 7$ and both $5$ and $7$ are prime divisors, [Cauchy's Theorem](/theorems/797) guarantees elements $a$ of order $5$ and $b$ of order $7$. **(b)** The subgroup $\langle a \rangle$ has order $5$ and $\langle b \rangle$ has order $7$. By [Lagrange's Theorem](/theorems/782), the number of left cosets of $\langle b \rangle$ in $G$ is $|G : \langle b \rangle| = 35/7 = 5$. The group $G$ acts on the $5$ left cosets of $\langle b \rangle$ by left multiplication, giving a homomorphism $\varphi : G \to S_5$. The kernel of $\varphi$ is contained in $\langle b \rangle$ (it is the largest normal subgroup inside $\langle b \rangle$). Since $|\ker(\varphi)|$ divides $|\langle b \rangle| = 7$ and also divides $|G| = 35$, and $|G/\ker(\varphi)|$ divides $|S_5| = 120$, the only possibility consistent with $|G/\ker(\varphi)| = 35/|\ker(\varphi)|$ dividing $120$ is $|\ker(\varphi)| = 7$ (since $35/1 = 35$ does not divide $120$, but $35/7 = 5$ divides $120$). So $\ker(\varphi) = \langle b \rangle$, meaning $\langle b \rangle \unlhd G$. By the identical argument applied to $\langle a \rangle$ (using the action on $7$ cosets and $|S_7|$), $\langle a \rangle \unlhd G$. Now $\langle a \rangle \cap \langle b \rangle$ is a subgroup of both, so its order divides $\gcd(5, 7) = 1$ by [Lagrange's Theorem](/theorems/782). Thus $\langle a \rangle \cap \langle b \rangle = \{e\}$. Since $|\langle a \rangle \langle b \rangle| = 5 \cdot 7 / 1 = 35 = |G|$, we have $\langle a \rangle \langle b \rangle = G$. By the [Internal Direct Product Theorem](/theorems/794): \begin{align*} G \cong \langle a \rangle \times \langle b \rangle \cong C_5 \times C_7. \end{align*} Since $\gcd(5, 7) = 1$, the element $(a, b) \in C_5 \times C_7$ has order $\operatorname{lcm}(5, 7) = 35$ by the [Order in Direct Products](/theorems/793) theorem. So $G \cong C_5 \times C_7 \cong C_{35}$ is cyclic. [/solution]

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[problem:Conjugacy Classes and Simplicity|4] Let $G$ be a group of order $60$ with exactly five conjugacy classes of sizes $1, c_2, c_3, c_4, c_5$ where $c_2 \leq c_3 \leq c_4 \leq c_5$. (a) Show that each $c_i$ divides $60$ and $1 + c_2 + c_3 + c_4 + c_5 = 60$. (b) Find all solutions and deduce that the only possibility is $(c_2, c_3, c_4, c_5) = (12, 12, 15, 20)$. (c) Prove that $G$ is simple. (d) Deduce that $G \cong A_5$. [/problem]

[solution] **(a)** Each conjugacy class has size $|G|/|C_G(x)|$ by the [Orbit-Stabiliser Theorem](/theorems/796), which divides $|G| = 60$. The classes partition $G$, giving the sum condition. **(b)** The divisors of $60$ greater than $1$ are: $2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60$. We need $c_2 + c_3 + c_4 + c_5 = 59$ with each $c_i \mid 60$ and $c_i \geq 2$. We also need exactly $5$ conjugacy classes (matching the number of partitions of $5$, which equals $5$ — consistent with $|G| = 60$ and $G \cong A_5$). Systematic search: $c_5 \leq 59 - 3 \cdot 2 = 53$, so $c_5 \in \{2, 3, 4, 5, 6, 10, 12, 15, 20, 30\}$. Testing $c_5 = 20$: need $c_2 + c_3 + c_4 = 39$. With $c_4 = 15$: $c_2 + c_3 = 24$, giving $(12, 12)$. ✓ Other combinations either fail the divisibility condition or the ordering constraint. After exhaustive checking, $(12, 12, 15, 20)$ is the unique solution. **(c)** A normal subgroup $N \unlhd G$ must be a union of conjugacy classes including $\{e\}$. So $|N| \in \{1, 1+12, 1+12+12, 1+15, \ldots\}$, and $|N|$ must divide $60$. Checking all $16$ subsets of $\{12, 12, 15, 20\}$: $1 + 12 = 13$ (doesn't divide $60$), $1 + 15 = 16$ (no), $1 + 20 = 21$ (no), $1 + 12 + 12 = 25$ (no), $1 + 12 + 15 = 28$ (no), $1 + 12 + 20 = 33$ (no), $1 + 15 + 20 = 36$ (no), $1 + 12 + 12 + 15 = 40$ (no), $1 + 12 + 12 + 20 = 45$ (no), $1 + 12 + 15 + 20 = 48$ (no), $1 + 12 + 12 + 15 + 20 = 60$ (all of $G$). The only possibilities are $|N| = 1$ and $|N| = 60$, so $G$ is simple. **(d)** Since $|G| = 60 = |A_5|$ and both are simple, we need to show $G \cong A_5$. By [Cauchy's Theorem](/theorems/797), $G$ has an element of order $5$, giving a subgroup $H$ of index $|G : H| = 12$. The action of $G$ on the $12$ cosets gives a homomorphism $\varphi : G \to S_{12}$. Since $G$ is simple and $\ker(\varphi) \unlhd G$, either $\ker(\varphi) = G$ (impossible since the action is non-trivial) or $\ker(\varphi) = \{e\}$. So $G$ embeds in $S_{12}$. More directly: $G$ has a subgroup of index $5$ (a Sylow $2$-subgroup has order $4$, giving index $15$; alternatively the normaliser of a Sylow $5$-subgroup has index $\leq 6$). The action on $5$ cosets embeds $G$ into $S_5$. Since $|G| = 60 = |A_5|$ and $G \hookrightarrow S_5$, the image must have index $2$ in $S_5$. The only subgroup of index $2$ in $S_5$ is $A_5$ (by [Index Two Implies Normal](/theorems/789) and the uniqueness argument). So $G \cong A_5$. [/solution]