September 22nd, 2026

1. We first verify the mappings are well-defined and mutual inverses. We know the image and preimage of a subgroup under a homomorphism are always subgroups. Furthermore, for any subgroup $H' \subseteq G'$, note $\phi(K) = \{\id_{G'}\} \subseteq H'$, meaning $K \subseteq \phi^{-1}(H')$. Thus the domains and codomains of the bijection are correct.

   To show they are inverses, note that $\phi(\phi^{-1}(H')) = H'$ follows immediately from the surjectivity of $\phi$. Conversely, we show $\phi^{-1}(\phi(H)) = H$ for any subgroup $H$ containing $K$. The inclusion $H \subseteq \phi^{-1}(\phi(H))$ is trivial. Let $x \in \phi^{-1}(\phi(H))$. Then $\phi(x) = \phi(h)$ for some $h \in H$. Rewriting yields $\phi(h^{-1}x) = \id_{G'}$, meaning $h^{-1}x \in K$. Because $K \subseteq H$, we have $h^{-1}x \in H$, and since $h \in H$, closure forces $x \in H$. Thus $\phi^{-1}(\phi(H)) \subseteq H$, meaning $\phi^{-1}(\phi(H)) = H$, establishing the bijection.

2. Suppose $H = \phi^{-1}(H')$ and $H' = \phi(H)$.

   • We previously proved that the preimage of a normal subgroup is always normal, and the image of a normal subgroup is normal provided the homomorphism is surjective. The result comes immediately.

   • The restriction $\phi|_H: H \to H'$ is inherently a homomorphism. It is surjective because $H' = \phi(H)$. The kernel of this restricted map is precisely $\{x \in H : \phi(x) = \id_{G'}\} = H \cap K$. Because $K \subseteq H$, this collapses to $K$, and so $\ker(\phi|_H) = K$.

   • By applying the standard counting formula to the restricted map $\phi|_H$, we immediately yield $|H| = |\ker(\phi|_H)| \times |\im(\phi|_H)| = |K| \times |H'|$.

We show that if $N$ is normal, the operation is strictly independent of the chosen representatives. Suppose $aN = a'N$ and $bN = b'N$. We wish to show that $(a'b')N = (ab)N$.

By the equivalence class properties of cosets, we can write $a' = a n_1$ and $b' = b n_2$ for some $n_1, n_2 \in N$. Then note

Because $N$ is normal, its left and right cosets coincide, meaning $Nb = bN$. Thus, for our element $n_1 \in N$, there exists some $n_3 \in N$ such that $n_1 b = b n_3$. Substituting this yields

Since $N$ is a subgroup and is closed under multiplication, $n_3n_2 \in N$. Thus $a'b'$ can be written as $ab$ multiplied by an element of $N$, which immediately implies $a'b' \in abN$. Consequently, $(a'b')N = (ab)N$.

If $N$ were not normal, the element $n_1 b$ could permanently escape $bN$, shifting the product into a completely different coset and causing the entire operation to be ill-defined. ❦

1. Closure. Since $a, b \in G$, their product $ab \in G$ by closure of the parent group, and so $(ab)N$ is indeed a valid coset in $G/N$.

2. Associativity. For any $aN, bN, cN \in G/N$, note that

   $$[(aN)(bN)](cN) = (ab)N \cdot cN = ((ab)c)N.$$

   Because the operation in $G$ is inherently associative, $((ab)c)N = (a(bc))N$. Thus

   $$(a(bc))N = aN \cdot (bc)N = (aN)[(bN)(cN)],$$

   satisfying associativity.

3. Identity. The coset of the identity, $\id_G N = N$, serves as the identity element. For any $aN \in G/N$, we have

   $$(aN)(N) = (a \id_G)N = aN, \quad \text{and} \quad (N)(aN) = (\id_G a)N = aN.$$

4. Inverse. The inverse of $aN$ is precisely $a^{-1}N$. Indeed,

   $$(aN)(a^{-1}N) = (aa^{-1})N = \id_G N = N,$$

   and symmetrically $(a^{-1}N)(aN) = N$.

First, we show $\pi$ is a homomorphism. For any $a, b \in G$, the definition of coset multiplication yields

and so $\pi$ preserves the group structure. Surjectivity is trivial, as any coset $aN \in G/N$ is immediately mapped to by $a \in G$. Finally, we determine the kernel. The identity element in $G/N$ is the coset $\id_G N = N$. Thus, $a \in \ker(\pi)$ if and only if $\pi(a) = N$, meaning $aN = N$. This occurs if and only if $a \in N$. Therefore, $\ker(\pi) = N$. ❦

We must establish both the existence and the uniqueness of $\tilde{f}$.

• Existence. We define $\tilde{f}$ directly on the equivalence classes in $A/{\sim}$ by setting $\tilde{f}([x]_\sim) = f(x)$. Because elements of $A/{\sim}$ are sets (equivalence classes) rather than individual elements of $A$, we must verify this function is well-defined. Suppose $[x]_\sim = [y]_\sim$. This implies $x \sim y$. Because we assumed $f$ respects the equivalence relation, $f(x) = f(y)$, and therefore $\tilde{f}([x]_\sim) = \tilde{f}([y]_\sim)$. The definition is independent of the chosen representative, so $\tilde{f}$ is a well-defined function. Furthermore, for any $x \in A$, we have $(\tilde{f} \circ \pi)(x) = \tilde{f}([x]_\sim) = f(x)$, ensuring the diagram commutes.

• Uniqueness. Suppose there is another function $g: A/{\sim} \to B$ such that $g \circ \pi = f$. We wish to show $g = \tilde{f}$. Let $[x]_\sim$ be an arbitrary element of $A/{\sim}$. Because $\pi$ is inherently surjective, $[x]_\sim = \pi(x)$ for some $x \in A$. Then $g([x]_\sim) = g(\pi(x)) = f(x)$. But we already know $f(x) = \tilde{f}(\pi(x)) = \tilde{f}([x]_\sim)$. Thus $g([x]_\sim) = \tilde{f}([x]_\sim)$ for all classes in the quotient, forcing $g = \tilde{f}$.

Since $(Q_1, \pi_1)$ satisfies the universal property and $\pi_2: G \to Q_2$ is a homomorphism with $\ker(\pi_2) = N \subseteq \ker(\pi_2)$ (trivially), there exists a unique homomorphism $\alpha: Q_1 \to Q_2$ such that $\alpha \circ \pi_1 = \pi_2$.

Symmetrically, since $(Q_2, \pi_2)$ satisfies the universal property and $\pi_1: G \to Q_1$ is a homomorphism with $N \subseteq \ker(\pi_1)$, there exists a unique homomorphism $\beta: Q_2 \to Q_1$ such that $\beta \circ \pi_2 = \pi_1$.

Substituting the first identity into the second yields $\beta \circ (\alpha \circ \pi_1) = \pi_1$, which we reassociate as $(\beta \circ \alpha) \circ \pi_1 = \pi_1$.

Now, consider the universal property of $Q_1$ applied to the homomorphism $\pi_1: G \to Q_1$. We seek a unique map $u: Q_1 \to Q_1$ such that $u \circ \pi_1 = \pi_1$. The identity map $\id_{Q_1}$ trivially satisfies this. However, we just showed that $\beta \circ \alpha$ also satisfies this. By the strict uniqueness guaranteed by the universal property, we are forced to conclude $\beta \circ \alpha = \id_{Q_1}$.

By an identical symmetric argument, $\alpha \circ \beta = \id_{Q_2}$. Thus $\alpha$ is a bijection, and since it is inherently a homomorphism, it is an isomorphism. This establishes that $Q_1 \cong Q_2$. ❦

Let $\phi: G \to H$ be a homomorphism such that $N \subseteq \ker(\phi)$. We must show there exists a unique homomorphism $\tilde{f}: G/N \to H$ making the diagram commute.

• Well-definedness. We propose $\tilde{f}(aN) = \phi(a)$. Because we are defining this on cosets (equivalence classes), we must check it is strictly independent of the representative. Suppose $aN = bN$. This means $b^{-1}a \in N$. Because $N \subseteq \ker(\phi)$, this implies $\phi(b^{-1}a) = \id_H$. Since $\phi$ is a homomorphism, $\phi(b)^{-1}\phi(a) = \id_H$, meaning $\phi(a) = \phi(b)$. Thus $\tilde{f}(aN) = \tilde{f}(bN)$, so $\tilde{f}$ is well-defined.

• Homomorphism. We verify that $\tilde{f}$ respects the group operation. Indeed, note

  $$\tilde{f}((aN)(bN)) = \tilde{f}((ab)N) = \phi(ab) = \phi(a)\phi(b) = \tilde{f}(aN)\tilde{f}(bN).$$

  Thus $\tilde{f}$ is a valid group homomorphism. By construction, $\tilde{f}(\pi(a)) = \tilde{f}(aN) = \phi(a)$, ensuring the diagram commutes.

• Uniqueness. Suppose $g: G/N \to H$ is another homomorphism satisfying $g \circ \pi = \phi$. For any coset $aN \in G/N$, we have $g(aN) = g(\pi(a)) = \phi(a)$. But $\phi(a) = \tilde{f}(aN)$. Since $g$ and $\tilde{f}$ agree on all elements, $g = \tilde{f}$.