- #1
Parmenides
- 37
- 0
Im asked to show that, given the groups [itex]H, G_1[/itex], and [itex]G_2[/itex] in which [itex]G_1 \cong G_2[/itex], that [tex]H\times{G_1} \cong H\times{G_2}[/tex]
Because of the isomorphism between [itex]G_1[/itex] and [itex]G_2[/itex], their cardinalities (order) are equal, which i think will be of good use when considering their Cartesian product with [itex]H[/itex]. So conceptually, it seems intuitive to believe that, almost vacuously, [tex]|H\times{G_1}| = |H\times{G_2}|[/tex] But I'm not sure how to explicitly show this. Since [itex]G_1 \cong{G_2}[/itex], there exists an isomorphism between the two such that: [tex]f: G_1 \rightarrow G_2[/tex] and that it is necessary to find an [itex]F[/itex] such that [tex]F: H\times{G_1} \rightarrow H \times G_2[/tex] Also, [itex]f(a) = b \forall a \in G_1, b \in G_2[/itex]. I think these are some pieces to the puzzle, but how to stitch them together?
Because of the isomorphism between [itex]G_1[/itex] and [itex]G_2[/itex], their cardinalities (order) are equal, which i think will be of good use when considering their Cartesian product with [itex]H[/itex]. So conceptually, it seems intuitive to believe that, almost vacuously, [tex]|H\times{G_1}| = |H\times{G_2}|[/tex] But I'm not sure how to explicitly show this. Since [itex]G_1 \cong{G_2}[/itex], there exists an isomorphism between the two such that: [tex]f: G_1 \rightarrow G_2[/tex] and that it is necessary to find an [itex]F[/itex] such that [tex]F: H\times{G_1} \rightarrow H \times G_2[/tex] Also, [itex]f(a) = b \forall a \in G_1, b \in G_2[/itex]. I think these are some pieces to the puzzle, but how to stitch them together?