[problem:Subgroups Of The Dihedral Group D4 | 2] Determine all subgroups of the dihedral group $D_4$ (the symmetry group of the square, of order $8$). For each subgroup, state whether it is normal in $D_4$. [/problem]

[solution] **Step 1 (List the elements).** The dihedral group $D_4$ has eight elements: four rotations $\{e, r, r^2, r^3\}$ (where $r$ is rotation by $90^\circ$) and four reflections $\{s, rs, r^2 s, r^3 s\}$, subject to the relations $r^4 = e$, $s^2 = e$, and $srs = r^{-1}$. **Step 2 (Subgroups of order 1 and 8).** The trivial subgroup $\{e\}$ and the full group $D_4$ are always subgroups, both normal. **Step 3 (Subgroups of order 2).** By [Lagrange's theorem](/theorems/841), any subgroup of order $2$ is cyclic, generated by an element of order $2$. The elements of order $2$ are $r^2$, $s$, $rs$, $r^2 s$, and $r^3 s$. This gives five subgroups of order $2$: \begin{align*} \{e, r^2\}, \quad \{e, s\}, \quad \{e, rs\}, \quad \{e, r^2 s\}, \quad \{e, r^3 s\}. \end{align*} To test normality, we check whether $g H g^{-1} = H$ for all $g \in D_4$. The subgroup $\{e, r^2\}$ is normal (since $r^2$ commutes with all rotations, and $s r^2 s^{-1} = s r^2 s = r^{-2} = r^2$). The four reflection subgroups are not normal: for instance, $r \{e, s\} r^{-1} = \{e, r s r^{-1}\} = \{e, r^2 s\} \neq \{e, s\}$. **Step 4 (Subgroups of order 4).** By Lagrange, subgroups of order $4$ have index $2$ in $D_4$, so they are automatically normal. The candidates are: - $\langle r \rangle = \{e, r, r^2, r^3\}$ — the rotation subgroup, cyclic of order $4$. - $\{e, r^2, s, r^2 s\}$ — isomorphic to the Klein four-group $V_4$, containing the rotation $r^2$ and two "opposite" reflections. - $\{e, r^2, rs, r^3 s\}$ — also isomorphic to $V_4$, containing $r^2$ and the other two reflections. All three are normal since they have index $2$. **Step 5 (Summary).** $D_4$ has exactly ten subgroups: one of order $1$, five of order $2$, three of order $4$, and one of order $8$. The normal subgroups are $\{e\}$, $\{e, r^2\}$, $\langle r \rangle$, $\{e, r^2, s, r^2 s\}$, $\{e, r^2, rs, r^3 s\}$, and $D_4$ itself — six in total. The three non-normal subgroups are the order-$2$ subgroups generated by individual reflections (not passing through $r^2$). [/solution]

[problem:Applying The First Isomorphism Theorem | 2] Let $\phi : \mathbb{Z} \to \mathbb{Z}/n\mathbb{Z}$ be the reduction map $\phi(k) = k \bmod n$. Verify that $\phi$ is a surjective group homomorphism, compute $\ker(\phi)$, and use the [first isomorphism theorem](/theorems/842) to conclude that $\mathbb{Z}/n\mathbb{Z} \cong \mathbb{Z}/n\mathbb{Z}$ (where the left-hand side is the quotient group and the right-hand side is the group of residue classes). [/problem]

[solution] **Step 1 (Homomorphism).** For $a, b \in \mathbb{Z}$, we have $\phi(a + b) = (a + b) \bmod n$. By the definition of modular addition, $(a + b) \bmod n = ((a \bmod n) + (b \bmod n)) \bmod n = \phi(a) + \phi(b)$ in $\mathbb{Z}/n\mathbb{Z}$. Thus $\phi$ preserves the group operation. **Step 2 (Surjectivity).** Every element $\bar{k} \in \mathbb{Z}/n\mathbb{Z}$ (for $0 \leq k \leq n - 1$) satisfies $\phi(k) = \bar{k}$, so $\phi$ is surjective. **Step 3 (Kernel).** An integer $k$ lies in $\ker(\phi)$ if and only if $k \bmod n = 0$, i.e., $n \mid k$. Therefore $\ker(\phi) = n\mathbb{Z}$. **Step 4 (Conclusion).** By the [first isomorphism theorem](/theorems/842), $\mathbb{Z}/\ker(\phi) \cong \operatorname{im}(\phi)$, i.e., $\mathbb{Z}/n\mathbb{Z} \cong \mathbb{Z}/n\mathbb{Z}$. This confirms that the abstract quotient group $\mathbb{Z}/n\mathbb{Z}$ (cosets of $n\mathbb{Z}$) is isomorphic to the concrete group of residue classes with modular addition. [/solution]

[problem:Orbit Counting With Burnside | 3] How many distinct necklaces can be made from $4$ beads, each of which is coloured either red or blue? Two necklaces are considered identical if one can be obtained from the other by rotation (but not reflection). Use [Burnside's lemma](/page/Burnside%27s%20Lemma): the number of distinct necklaces is \begin{align*} \frac{1}{|G|} \sum_{g \in G} |\operatorname{Fix}(g)|, \end{align*} where $G$ is the group of rotations acting on colourings and $\operatorname{Fix}(g)$ is the number of colourings fixed by $g$. [/problem]

[solution] **Step 1 (Identify the group and the set).** The group is $G = \mathbb{Z}/4\mathbb{Z} = \{r^0, r^1, r^2, r^3\}$, the cyclic group of rotations of a square, acting on the set $X$ of all $2^4 = 16$ colourings of $4$ beads with $2$ colours. **Step 2 (Count fixed points for each group element).** *$r^0$ (identity):* Every colouring is fixed. $|\operatorname{Fix}(r^0)| = 16$. *$r^1$ (rotation by $90^\circ$):* A colouring is fixed if and only if all four beads have the same colour. $|\operatorname{Fix}(r^1)| = 2$ (all red or all blue). *$r^2$ (rotation by $180^\circ$):* A colouring is fixed if and only if opposite beads have the same colour. There are two pairs of opposite beads, each pair can be coloured independently in $2$ ways. $|\operatorname{Fix}(r^2)| = 2^2 = 4$. *$r^3$ (rotation by $270^\circ$):* By the same argument as $r^1$, all four beads must be the same colour. $|\operatorname{Fix}(r^3)| = 2$. **Step 3 (Apply Burnside's lemma).** By Burnside's lemma, the number of distinct necklaces is: \begin{align*} \frac{1}{|G|} \sum_{g \in G} |\operatorname{Fix}(g)| = \frac{1}{4}(16 + 2 + 4 + 2) = \frac{24}{4} = 6. \end{align*} **Step 4 (Verification).** The six necklaces are: RRRR, BRRR, BBRR, BRBR, BBBR, BBBB (listing one representative from each orbit). Without the group-theoretic framework, verifying that these are exactly the orbits requires checking each of the $16$ colourings against all others — Burnside's lemma replaces this with a single computation. [/solution]