Let \(\phi:G_1\to G_2\) be a group homomorphism and let \(H_2\le G_2\). Put \[ H_1=\phi^{-1}(H_2). \]
- Show that \(\phi(G_1)\le G_2\) and \(H_1\le G_1\).
- Show that \(\ker\phi\trianglelefteq G_1\).
- Prove \[ xH_1=yH_1 \iff \phi(x)H_2=\phi(y)H_2. \]
- Show that \(\phi\) is injective iff \(\ker\phi=\{e\}\).
- Prove the first isomorphism theorem \[ G_1/\ker\phi\cong\operatorname{im}\phi. \]
Solution
1. Images and preimages are subgroups.
If \(a=\phi(x)\) and \(b=\phi(y)\) lie in \(\phi(G_1)\), then \[ ab^{-1}=\phi(xy^{-1})\in\phi(G_1), \] so the image is a subgroup.
If \(x,y\in H_1\), then \(\phi(x),\phi(y)\in H_2\), hence \[ \phi(xy^{-1})=\phi(x)\phi(y)^{-1}\in H_2. \] Thus \(xy^{-1}\in H_1\).
2. The kernel is normal.
If \(k\in\ker\phi\) and \(g\in G_1\), then \[ \phi(gkg^{-1})=\phi(g)e\phi(g)^{-1}=e, \] so \(gkg^{-1}\in\ker\phi\).
3. The coset criterion holds.
We have \[ xH_1=yH_1 \iff y^{-1}x\in H_1 \iff \phi(y)^{-1}\phi(x)\in H_2 \iff \phi(x)H_2=\phi(y)H_2. \]
4. Injectivity is equivalent to trivial kernel.
If \(\phi\) is injective, then \(\phi(g)=e\) implies \(g=e\). Conversely, if \(\ker\phi=\{e\}\) and \(\phi(x)=\phi(y)\), then \[ \phi(y^{-1}x)=e, \] so \(y^{-1}x=e\) and \(x=y\).
5. First isomorphism theorem.
4. Hence it is an isomorphism.