Normal Subgroups and Quotient Groups

In $D_4 = \{e, r, r^2, r^3, s, rs, r^2s, r^3s\}$ (Topic 6), let $N = \langle r\rangle = \{e, r, r^2, r^3\}$, the rotations. Since $N$ already contains every power of $r$, conjugating by a rotation keeps $N$ fixed pointwise. Conjugating by the reflection $s$ sends $r$ to $srs^{-1} = srs = r^{-1} = r^3 \in N$ (using $sr = r^{-1}s$ and $s^2 = e$), and every other element of $N$ along with it. So $sNs^{-1} = N$ as well, and $N \trianglelefteq D_4$.

Let $H = \langle s\rangle = \{e, s\}$, order $2$, and test conjugation by $r$: $rsr^{-1} = rsr^3$. Using $sr = r^{-1}s$ repeatedly, $rsr^3 = (rs)r^3 = r(sr)r^2 = r(r^{-1}s)r^2 = sr^2 = r^2s$ (moving $s$ past each $r$ flips its sign). So $rsr^{-1} = r^2s$, which is neither $e$ nor $s$: $r^2s \notin H$. One failing conjugate is enough, so $H$ is not normal in $D_4$, even though $|H|$ divides $|D_4|$ and $H$ passes Lagrange's theorem fine.

Let $N = \{e, (1\,2)(3\,4), (1\,3)(2\,4), (1\,4)(2\,3)\}$, the identity together with the three ways to pair up $\{1,2,3,4\}$ into two disjoint $2$-cycles. First, $N$ is a subgroup: any two of the three double-transpositions compose to the third or to $e$ (for instance $(1\,2)(3\,4)\cdot(1\,3)(2\,4) = (1\,4)(2\,3)$). For normality, conjugating a double-transposition by any $g \in S_4$ just relabels its two pairs by $g$, which can only produce another double-transposition (or, if $g$ itself fixes the pairing, the same one back) — never a $3$-cycle or a single transposition. So conjugation always lands back inside $N$, and $N \trianglelefteq S_4$.

Let $N = \langle 6\rangle = \{0, 6\}$ in $\mathbb Z_{12}$, order $2$, so $[{\mathbb Z_{12}}:N] = 12/2 = 6$. The six cosets are $k + N$ for $k = 0, \ldots, 5$. Adding cosets by adding representatives, $(1+N)+(1+N) = 2+N$, and repeating, $k\cdot(1+N) = k+N$, which only returns to $N$ when $6 \mid k$. So $1+N$ has order $6$ in the quotient, generating all of it: $\mathbb Z_{12}/\langle 6\rangle \cong \mathbb Z_6$.

Let $N = A_3 = \{e, (1\,2\,3), (1\,3\,2)\} \trianglelefteq S_3$ (normal since it is the kernel of the sign map, Topic 6). There are $[S_3:A_3] = 6/3 = 2$ cosets, $N$ and $(1\,2)N$. Multiplying the nontrivial coset by itself: $(1\,2)N \cdot (1\,2)N = (1\,2)(1\,2)N = eN = N$, since $(1\,2)^2 = e$. So $S_3/A_3$ has two elements, one of which squares to the identity — it is $\mathbb Z_2$, even though $S_3$ itself is non-abelian.

Let $N = \langle r^2\rangle = \{e, r^2\} \trianglelefteq D_4$ (normal: $r^2$ is fixed by conjugation by $r$, and $srs^{-1}=r^{-1}=r^3$ also squares to $r^2$, more carefully checked in Why It Works). There are $8/2 = 4$ cosets: $N, rN, sN, rsN$. Since $r^2 \in N$ and $s^2 = e \in N$, both $(rN)^2 = r^2N = N$ and $(sN)^2 = N$: every generator has order $2$ in the quotient. Checking the mixed product both ways, $(rN)(sN) = rsN$ and $(sN)(rN) = srN = r^{-1}sN = r^3sN$; since $r^3s = r^2\cdot rs$ differs from $rs$ by the element $r^2 \in N$, $r^3sN = rsN$ as cosets — so the two orders agree and $D_4/N$ is abelian. With three elements of order $2$ besides the identity, $D_4/\langle r^2\rangle \cong \mathbb Z_2 \times \mathbb Z_2$.

Let $\varphi: \mathbb Z_{12} \to \mathbb Z_4$ send $k \mapsto k \bmod 4$. Then $\ker\varphi = \{0, 4, 8\}$ (order $3$) and $\operatorname{im}\varphi = \mathbb Z_4$ (every residue is hit, order $4$). The First Isomorphism Theorem predicts $\mathbb Z_{12}/\{0,4,8\} \cong \mathbb Z_4$, and indeed the quotient has order $12/3 = 4$, matching $|\operatorname{im}\varphi|$.

For $\operatorname{sgn}: S_4 \to \{1,-1\}$, $\ker(\operatorname{sgn}) = A_4$, the even permutations, with $|A_4| = 12$ (half of $|S_4|=24$, since the even and odd permutations are in bijection via multiplying by any fixed transposition). The image is all of $\{1,-1\}$, order $2$. First Isomorphism Theorem: $S_4/A_4 \cong \mathbb Z_2$, and $24/12 = 2$ checks out.

Let $\varphi: U(16) \to U(16)$ send $x \mapsto x^2 \bmod 16$; it is a homomorphism since $U(16) = \{1,3,5,7,9,11,13,15\}$ is abelian, so $(xy)^2 = x^2y^2$. Squaring every element: $1,9,9,1,1,9,9,1$ for $x=1,3,5,7,9,11,13,15$ respectively (e.g. $5^2=25\equiv9$, $7^2=49\equiv1 \pmod{16}$). So $\ker\varphi = \{1,7,9,15\}$, order $4$, and $\operatorname{im}\varphi = \{1,9\}$, order $2$; indeed $4 \cdot 2 = 8 = |U(16)|$. By the theorem, $U(16)/\{1,7,9,15\} \cong \{1,9\} \cong \mathbb Z_2$.

Let $H = \langle 4\rangle = \{0,4,8,12,16,20\}$ (order $6$) and $N = \langle 6\rangle = \{0,6,12,18\}$ (order $4$) in $\mathbb Z_{24}$; both are normal since $\mathbb Z_{24}$ is abelian. Then $H \cap N = \langle 12\rangle = \{0,12\}$ (order $2$, generated by $\operatorname{lcm}(4,6)=12$) and $HN = \langle 2\rangle$, all $12$ even residues (generated by $\gcd(4,6)=2$). Checking the theorem: $|HN|/|N| = 12/4 = 3$ and $|H|/|H\cap N| = 6/2 = 3$ — they agree, so $HN/N \cong H/(H\cap N) \cong \mathbb Z_3$.

Keep $K = \langle 4\rangle = \{0,4,8,12,16,20\}$ (order $6$) and let $N = \langle 12\rangle = \{0,12\}$ (order $2$); since $12$ is a multiple of $4$, $N \le K$. Then $|G/N| = 24/2 = 12$ and $|K/N| = 6/2 = 3$, so $(G/N)/(K/N)$ has order $12/3 = 4$. Directly, $|G/K| = 24/6 = 4$ as well — the two sides match, both isomorphic to $\mathbb Z_4$, exactly as $(G/N)/(K/N) \cong G/K$ demands.

In $S_4$, let $N = \{e,(1\,2)(3\,4),(1\,3)(2\,4),(1\,4)(2\,3)\} \trianglelefteq S_4$ (shown normal above) and $H = \langle(1\,2\,3)\rangle = \{e,(1\,2\,3),(1\,3\,2)\}$, order $3$ — itself NOT normal in $S_4$, since conjugating $(1\,2\,3)$ by $(1\,4)$ gives $(4\,2\,3) = (2\,3\,4) \notin H$. Since $H$ has order $3$ (prime) and $N$ has only elements of order $1$ or $2$, $H \cap N = \{e\}$. Then $HN$ has order $3\cdot4/1 = 12$; since $H, N \subseteq A_4$ and $|A_4|=12$ (Topic 5), $HN = A_4$. The theorem gives $A_4/N \cong H/\{e\} \cong H \cong \mathbb Z_3$ — a clean isomorphism, even though $H$ itself was never normal in $S_4$.

From Section 2, $r^2$ commutes with every rotation (rotations commute with each other) and with every reflection, since $sr^2s^{-1} = (srs^{-1})^2 = (r^{-1})^2 = r^{-2} = r^2$ — so $r^2 \in Z(D_4)$. No other nonidentity element works: $r \notin Z(D_4)$ since $srs^{-1}=r^{-1}\ne r$, and no reflection commutes with $r$ either, by the same computation that showed $\langle s\rangle$ is not normal. So $Z(D_4) = \{e, r^2\}$ exactly — the same subgroup by which Section 2 found the abelian quotient $D_4/\langle r^2\rangle \cong \mathbb Z_2\times\mathbb Z_2$.

Test $r=(1\,2\,3)$: it fails to commute with $s=(1\,2)$, since $rs = (1\,2\,3)(1\,2) = (1\,3)$ while $sr = (1\,2)(1\,2\,3) = (2\,3)$, and $(1\,3)\ne(2\,3)$. So $r \notin Z(S_3)$, and by symmetry neither does $r^2$ or any transposition (each fails against some other element the same way). With every nonidentity element ruled out, $Z(S_3) = \{e\}$: $S_3$ has a trivial center, even though some of its elements (like $r$) commute with SOME others (like $r^2$).

Since $S_4/A_4 \cong \mathbb Z_2$ is abelian (shown in Section 3), $A_4$ is a normal subgroup with abelian quotient, so $S_4' \subseteq A_4$ — the commutator subgroup can never be bigger than any such subgroup. For the reverse inclusion, compute one commutator: $a=(1\,2)$, $b=(1\,3)$ give $[a,b] = a^{-1}b^{-1}ab = (1\,2)(1\,3)(1\,2)(1\,3) = (1\,2\,3)$, a $3$-cycle. Since $3$-cycles generate all of $A_4$ (Topic 4), and $S_4'$ is a subgroup containing this $3$-cycle, $S_4' \supseteq A_4$ as well. Together, $S_4' = A_4$, order $12$ — while $Z(S_4) = \{e\}$ is trivial, the same way $Z(S_3)$ was: the center and the commutator subgroup can be wildly different sizes.