Isomorphism Theorems

1st isomorphism theorem: if $f : G \rightarrow H$ is a homomorphism of groups, then ker(f) $\unlhd G$ and $G/ker(f) \cong f(G)$.

Proof: Let $\phi : G/ker(f) \rightarrow im(f)$ be a homomorphism that sends x ker(f) $\mapsto$ f(n). This is a well defined map because x ker(f) = y ker(y) $\rightarrow$ f(x) = f(y). Let $x^{-1}y \in$ ker(f), then $f(x^{-1}y) = 1 \rightarrow f(x)^{-1}f(y) \rightarrow f(x) = f(y)$. It is a well defined homomorphism because $\phi$(x ker(f) $\cdot$ y ker(y)) = $\phi$(xy ker(f)) = $\phi(xy) = \phi(x)\phi(y) = \phi$(x ker(f))$\phi$(y ker(f)). Let $y \in im(f)$. Then y = f(n) = $\phi$(n ker(n)) so it is surjective. Note that ker$(\phi) = \{ \text{ x ker(f)} \vert f(n) = 1 \} = \{ \text{x ker(f)} : x \in ker(f)\} = \{\text{ker(f)}\}$ so $\phi$ is injective. This completes the proof.

Remark: the size of the image is $\vert Im(f) \vert = [G : ker(f)]$. If G is finite, $\vert Im(f) \vert = \frac{\vert G \vert}{\vert ker(f) \vert}$ and $\vert G \vert = \vert Im(f) \vert \vert ker(f) \vert$.

2nd isomorphism theorem: let $H, K \leq G$ and $H \leq N_G(K)$, then $K \unlhd HK \leq G, H \cap K \unlhd H$ and $HK/K \cong H/H\cap K$

Proof: let $\phi : HK \rightarrow H/H \cap K$ be a map defined by $x \cdot y \mapsto x(H \cap K)$. It is well defined because if $x_1 \cdot y_1 = x_2 \cdot y_2$ then $x_2^{-1}\cdot x_1 = y_2 \cdot y_1^{-1} \in H \cap K$ so $x_1H\cap K = x^2 H \cap K$. It is a homomorphism because $\phi(x_1 \cdot y_1 \cdot x_2 \cdot y_2) =$ $\phi(x_1x_2x_2^{-1}y_1x_1y_2) = x_1x_2(H \cap K) =$ $x_1(H \cap K)x_2(H \cap K) =$ $\phi(x_1y_1) \cdot \phi(x_2y_2)$. By definition HK is a subgroup, and we know that $H \cap K \leq H$. To show that it is normal see that if $x \in H \cap K, y \in H, yxy^{-1} \in H$ and $yxy^{-1} \in K$ hence $yxy^{-1} \in H \cap K$ and it is normal. Im($\phi$) = $H/H \cap K, xH\cap K = \phi(x \cdot 1)$, ker($\phi) = \{ x \cdot y : x(H \cap K) = H \cap K \} (x \in H \cap K$ so K = ker($\phi$), which means K is normal. By the first isomorphism theorem $Im(\phi) = H/H \cap K \cong HK/K$ (here K = ker($\phi$)), completing the proof.

Note that if $H \unlhd HK$ then we would have $HK/H \cong K/H \cap K$.

3rd isomorphism theorem: let G be a group and let H and K be normal subgroups of G with $H \leq K$. Then $K/H \unlhd G/H$ and $(G/H)(K/H) \cong G/K$.

Proof: the proof that $K/H \unlhd G/H$ is left as an exercise to the reader. Let $\phi : G/H \rightarrow G/K$ be a map defined by $(gH) \mapsto gK$. Assume $g_1H = g_2H$ then $g_1 = g_2h$ for some $h \in H$. Since $H \leq K, h \in K$ so $g_1K = g_2K$ which means $\phi(g_1H) = \phi(g_2H)$ so it is well defined. ker$(\phi) = \{gH \in G/H \vert \phi(gH) =$ $1K\} =$ $\{gH \in G/H \vert gK = 1K\} =$ $\{gH \in G/H \vert g \in K \} =$ $K/H.$ By the first isomorphism theorem $(G/H)(K/H) \cong G/K$