I'm trying to figure out when and which isomorphism theorems are used in the following proof, as well as if there is another way of proving that relies more on the theorems:

Let $A\triangleleft G$ and $B\triangleleft H$. Show that $(A\times B)\triangleleft (G\times H)$ and $$(G\times H)/(A\times B)\cong (G/A)\times (H/B).$$


Define the homomorphism $\varphi: G\times H \rightarrow (G/A)\times (H/B)$ such that $\varphi((g,h))=(gA,hB).$ This is a homomorphism because for $g_1,g_2\in G, h_1,h_2\in H$ we have $\varphi ((g_1,h_1),(g_2,h_2))=\varphi((g_1g_2,h_1h_2))=(g_1g_2A,h_1h_2B)=(g_1G,h_1B)(g_2A,h_2B)=\varphi((g_1,h_1))\varphi((g_2,h_2)).$ The kernal of $\varphi$ is any element $(g,h)$ such that $g\in A$ and $h\in B$. This yields $A\times B$ therefore, $(A\times B) \triangleleft (G\times H$) and ($G\times H)/(A\times B)\cong (G/A)\times (H/B).$


1 Answer 1


One only uses the first isomorphism theorem of groups which asserts that

$$G/\operatorname{ker f} \cong f(G)$$

for a group morphism $f:G \to H$.

The proof finds the kernel and then applies this theorem. Note that the map in the proof is trivially surjective, so the image is equal to the codomain


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