The Isomorphism Theorems

We give complete, self-contained proofs, each building only on the previous ones and on Quotient Groups and Group Homomorphisms. Throughout, $e$ denotes the identity of whichever group is meant by context.

Proof of the First Isomorphism Theorem. Step 1: $\ker\varphi \trianglelefteq G$. First, $\ker\varphi$ is a subgroup: $\varphi(e) = e$ (since $\varphi(e) = \varphi(ee) = \varphi(e)\varphi(e)$, and canceling $\varphi(e)$ in $H$ gives $\varphi(e) = e$), so $e \in \ker\varphi$. If $a, b \in \ker\varphi$ then $\varphi(ab^{-1}) = \varphi(a)\varphi(b)^{-1} = e\cdot e^{-1} = e$, using that $\varphi(b^{-1}) = \varphi(b)^{-1}$ (which follows from $\varphi(b)\varphi(b^{-1}) = \varphi(bb^{-1}) = \varphi(e) = e$). So $\ker\varphi$ is a subgroup by the subgroup criterion. For normality, let $n \in \ker\varphi$ and $g \in G$. Then

so $gng^{-1} \in \ker\varphi$. Since $g$ and $n$ were arbitrary, $g(\ker\varphi)g^{-1} \subseteq \ker\varphi$ for every $g \in G$, which is the definition of normality. Step 2: $\operatorname{im}\varphi \le H$. Nonempty since $e = \varphi(e) \in \operatorname{im}\varphi$. If $\varphi(a), \varphi(b) \in \operatorname{im}\varphi$ then $\varphi(a)\varphi(b)^{-1} = \varphi(ab^{-1}) \in \operatorname{im}\varphi$. So $\operatorname{im}\varphi$ is a subgroup of $H$ by the subgroup criterion. Step 3: $\bar\varphi$ is well defined. Write $N = \ker\varphi$. Suppose $g_1 N = g_2 N$ in $G/N$; we must show $\varphi(g_1) = \varphi(g_2)$, since $\bar\varphi$ is defined on cosets by evaluating a representative. From $g_1 N = g_2N$ we get $g_2^{-1}g_1 \in N$, i.e. $\varphi(g_2^{-1}g_1) = e$. Then

so $\varphi(g_1) = \varphi(g_2)$. This shows $\bar\varphi$ does not depend on the choice of coset representative. Step 4: $\bar\varphi$ is a homomorphism. For any $g_1, g_2 \in G$,

using the definition of the quotient group operation (valid because $N \trianglelefteq G$ by Step 1) and the homomorphism property of $\varphi$. Step 5: $\bar\varphi$ is injective. Suppose $\bar\varphi(gN) = e$. Then $\varphi(g) = e$, so $g \in N = \ker\varphi$, so $gN = N$, the identity coset. Thus the kernel of $\bar\varphi$ is trivial, and a homomorphism with trivial kernel is injective: if $\bar\varphi(g_1N) = \bar\varphi(g_2N)$ then $\bar\varphi(g_1N (g_2N)^{-1}) = e$ (using Step 4 and that homomorphisms send inverses to inverses), so $g_1N(g_2N)^{-1} = N$, i.e. $g_1N = g_2N$. Step 6: $\bar\varphi$ is surjective onto $\operatorname{im}\varphi$. Every element of $\operatorname{im}\varphi$ has the form $\varphi(g) = \bar\varphi(gN)$ for some $g \in G$, by definition of image. Steps 3–6 show $\bar\varphi: G/N \to \operatorname{im}\varphi$ is a well-defined bijective homomorphism, i.e. an isomorphism. $\blacksquare$

Proof of the Second Isomorphism Theorem. Fix $H \le G$ and $N \trianglelefteq G$. Step 1: $HN \le G$. Since $e \in H$ and $e \in N$, $e = ee \in HN$, so $HN \ne \varnothing$. Let $h_1n_1, h_2n_2 \in HN$ with $h_i \in H$, $n_i \in N$. Then

Since $N \trianglelefteq G$, we have $h_2^{-1}(n_1n_2^{-1})h_2 \in N$ (conjugating an element of $N$ by $h_2 \in G$ stays in $N$); call this element $n_3 \in N$. Then

where $n_3' = h_2 n_1n_2^{-1}h_2^{-1} \in N$ by normality of $N$ (conjugation by $h_2$). Since $h_1h_2^{-1} \in H$ and $n_3' \in N$, we get $(h_1n_1)(h_2n_2)^{-1} \in HN$. By the subgroup criterion, $HN \le G$. Step 2: $N \trianglelefteq HN$. Certainly $N \le HN$ (take $h = e$), and $N \le G$ acts on itself with $HN \le G$, so for any $x \in HN \subseteq G$ and $n \in N$, $xnx^{-1} \in N$ because $N \trianglelefteq G$. This is precisely normality of $N$ inside the subgroup $HN$. Step 3: $H \cap N \trianglelefteq H$. $H \cap N$ is a subgroup as an intersection of two subgroups. For $h \in H$ and $x \in H \cap N$: $hxh^{-1} \in H$ since $H$ is a subgroup and $x, h \in H$; also $hxh^{-1} \in N$ since $N \trianglelefteq G$ and $x \in N$, $h \in G$. So $hxh^{-1} \in H \cap N$, proving normality in $H$. Step 4: the isomorphism. Define $\varphi: H \to HN/N$ by $\varphi(h) = hN$. This is well defined since $h \in H \subseteq HN$, so $hN$ is indeed a coset in $HN/N$ (using Step 2). It is a homomorphism because $\varphi(h_1h_2) = h_1h_2N = (h_1N)(h_2N) = \varphi(h_1)\varphi(h_2)$, the middle equality being the quotient group operation in $HN/N$. $\varphi$ is surjective: an arbitrary element of $HN/N$ is $(hn)N$ for some $h \in H, n\in N$, and $(hn)N = hN$ since $n \in N$; so $(hn)N = \varphi(h)$. Now compute $\ker\varphi$: $\varphi(h) = N$ (the identity of $HN/N$) iff $hN = N$ iff $h \in N$. Combined with $h \in H$, this means $\ker\varphi = H \cap N$. By the First Isomorphism Theorem applied to $\varphi: H \to HN/N$,

using surjectivity of $\varphi$ from above. This proves the theorem. $\blacksquare$

Proof of the Third Isomorphism Theorem. Let $N \le K$, both normal in $G$. Step 1: define the map. Let $\pi_K : G \to G/K$ be the canonical projection, a surjective homomorphism with kernel $K$ (this is immediate from the definition of the quotient group operation: $\pi_K$ is a homomorphism because $\pi_K(g_1g_2) = g_1g_2K = (g_1K)(g_2K)$, surjective by definition of cosets, and its kernel is $\{g : gK = K\} = K$). Since $N \le K = \ker \pi_K$, define

Step 2: $\psi$ is well defined. If $g_1N = g_2N$ then $g_2^{-1}g_1 \in N \subseteq K$, so $g_2^{-1}g_1 \in K$, hence $g_1K = g_2K$, i.e. $\psi(g_1N) = \psi(g_2N)$. (Here we use $N \le K$ essentially.) Step 3: $\psi$ is a surjective homomorphism. $\psi\big((g_1N)(g_2N)\big) = \psi(g_1g_2N) = g_1g_2K = (g_1K)(g_2K) = \psi(g_1N)\psi(g_2N)$, using the quotient operations in $G/N$ and $G/K$ (both valid since $N, K \trianglelefteq G$). Surjectivity: every coset $gK \in G/K$ equals $\psi(gN)$. Step 4: compute $\ker\psi$. $\psi(gN) = K$ (identity of $G/K$) iff $gK = K$ iff $g \in K$. The set of cosets $gN$ with $g \in K$ is exactly $K/N = \{kN : k \in K\}$, which is a well-defined subset of $G/N$ since $N \le K$. So $\ker\psi = K/N$. Since kernels of homomorphisms are always normal (proved in Step 1 of the First Isomorphism Theorem, applied to $\psi$), $K/N \trianglelefteq G/N$. Step 5: conclude. By the First Isomorphism Theorem applied to $\psi: G/N \to G/K$,

using Step 3 (surjectivity) and Step 4. $\blacksquare$

Proof of the Lattice Isomorphism Theorem. Let $N \trianglelefteq G$ and $\pi: G \to G/N$ the canonical projection. Step 1: the maps are well defined and mutually inverse. For $A \in \mathcal A$ (so $N \le A \le G$), let $f(A) = A/N = \{aN : a \in A\}$; since $N \trianglelefteq G$ and $N \le A$, $N \trianglelefteq A$ as well (the same computation as Step 3 of the Second Isomorphism Theorem, with $H$ replaced by $A$), so $A/N$ is a genuine group and a subset of $G/N$; one checks directly it's a subgroup of $G/N$ (closed under the quotient operation and inverses since $A$ is a subgroup of $G$). So $f: \mathcal A \to \mathcal B$ is well defined, landing in subgroups of $G/N$. For $B \in \mathcal{B}$, let $g(B) = \pi^{-1}(B) = \{x \in G : \pi(x) \in B\}$. This is a subgroup of $G$: it contains $e$ since $\pi(e) = N \in B$ (identity of $G/N$, and $B$ a subgroup); it's closed under products and inverses because $\pi$ is a homomorphism (if $\pi(x), \pi(y) \in B$ then $\pi(xy^{-1}) = \pi(x)\pi(y)^{-1} \in B$ since $B$ is a subgroup). Moreover $N \le \pi^{-1}(B)$ always, since $\pi(N) = \{N\} \subseteq B$ (every subgroup contains the identity). So $g: \mathcal B \to \mathcal A$ is well defined. We check $g(f(A)) = A$ for $A \in \mathcal A$: $g(f(A)) = \pi^{-1}(A/N) = \{x \in G : xN \in A/N\}$. If $x \in A$ then $xN \in A/N$ trivially, so $A \subseteq g(f(A))$. Conversely if $xN \in A/N$ then $xN = aN$ for some $a \in A$, so $x \in aN \subseteq A$ (since $a\in A$ and $N \le A$ means $aN \subseteq A$). So $g(f(A)) \subseteq A$. Hence $g(f(A)) = A$. We check $f(g(B)) = B$ for $B \in \mathcal B$: $f(g(B)) = \pi^{-1}(B)/N$. Since $\pi$ is surjective, $\pi(\pi^{-1}(B)) = B$ (standard fact for surjective maps: for any subset $B$ of the codomain, $\pi(\pi^{-1}(B)) = B \cap \operatorname{im}\pi = B$ as $\pi$ is onto). But $\pi(\pi^{-1}(B)) = \{xN : x \in \pi^{-1}(B)\} = \pi^{-1}(B)/N$ by definition of the quotient map on cosets. So $\pi^{-1}(B)/N = B$, i.e. $f(g(B)) = B$. This proves $f$ and $g$ are mutually inverse bijections between $\mathcal A$ and $\mathcal B$. Step 2: order preservation. If $A_1 \le A_2$ (both in $\mathcal A$), then clearly every coset $aN$ with $a \in A_1$ is a coset $aN$ with $a \in A_2$, so $A_1/N \subseteq A_2/N$, i.e. $B_1 \le B_2$. Conversely if $B_1 \le B_2$, apply $g$ (order-preserving by the same argument, since $\pi^{-1}$ preserves inclusions) to get $A_1 = g(B_1) \le g(B_2) = A_2$. So $A_1 \le A_2 \iff B_1 \le B_2$. For indices: define $\Phi: A_2/A_1 \to B_2/B_1$ by $\Phi(aA_1) = (aN)B_1$ for $a \in A_2$. This is well defined and bijective by an argument identical in structure to the Third Isomorphism Theorem's map $\psi$, applied to the restriction of $\pi$ to $A_2$ with kernel $N$ inside $A_2$, composed with identifying $A_1/N = B_1$: concretely, $\pi|_{A_2} : A_2 \to B_2$ is a surjective homomorphism with kernel $A_2 \cap N = N$ (since $N \le A_2$) and by the Third Isomorphism theorem argument, $A_2/A_1 \cong B_2/B_1$ via cosets, giving $[A_2:A_1] = |A_2/A_1| = |B_2/B_1| = [B_2:B_1]$. Step 3: normality preservation. Suppose $A_1 \trianglelefteq G$. For $\bar g = gN \in G/N$ and $\bar a = aN \in B_1 = A_1/N$, we have $\bar g \bar a \bar g^{-1} = (gag^{-1})N$. Since $A_1 \trianglelefteq G$, $gag^{-1} \in A_1$, so $(gag^{-1})N \in A_1/N = B_1$. As $\bar g, \bar a$ were arbitrary, $B_1 \trianglelefteq G/N$. Conversely suppose $B_1 \trianglelefteq G/N$. For $g \in G$ and $a \in A_1$, we have $\pi(gag^{-1}) = \pi(g)\pi(a)\pi(g)^{-1} \in B_1$ since $\pi(a) \in B_1$ and $B_1 \trianglelefteq G/N$. So $gag^{-1} \in \pi^{-1}(B_1) = A_1$ (using Step 1, $g(f(A_1)) = A_1$ i.e. $\pi^{-1}(B_1) = A_1$). As $g, a$ were arbitrary, $A_1 \trianglelefteq G$. When $A_1 \trianglelefteq G$ (equivalently $B_1 \trianglelefteq G/N$), apply the Third Isomorphism Theorem to the normal subgroups $N \le A_1$ of $G$: $(G/N)/(A_1/N) \cong G/A_1$, i.e. $(G/N)/B_1 \cong G/A_1$. Step 4: lattice operations. For intersections: $f(A_1 \cap A_2) = (A_1\cap A_2)/N$. We claim this equals $B_1 \cap B_2 = (A_1/N) \cap (A_2/N)$. If $x \in A_1 \cap A_2$ then $xN \in A_1/N$ and $xN \in A_2/N$, so $xN \in B_1 \cap B_2$; this gives $\subseteq$. Conversely if $xN \in B_1 \cap B_2$ then $xN = a_1N$ for some $a_1 \in A_1$ and $xN = a_2N$ for some $a_2 \in A_2$; since $x \in xN = a_1N \subseteq A_1$ (as $N \le A_1$) and likewise $x \in A_2$, we get $x \in A_1 \cap A_2$, so $xN \in f(A_1\cap A_2)$. This gives equality. For the join $\langle A_1, A_2\rangle$: since $f$ is order-preserving and $A_1, A_2 \le \langle A_1,A_2\rangle$, we get $B_1, B_2 \le f(\langle A_1,A_2\rangle)$, so $\langle B_1,B_2\rangle \le f(\langle A_1,A_2\rangle)$. Conversely, $g(\langle B_1,B_2\rangle) \ge g(B_1) = A_1$ and $\ge g(B_2) = A_2$ (order preservation of $g$), so $g(\langle B_1,B_2\rangle) \ge \langle A_1,A_2\rangle$, i.e. (applying $f$, order-preserving and a bijection) $\langle B_1,B_2\rangle \ge f(\langle A_1,A_2\rangle)$. Combining both inclusions gives equality. $\blacksquare$