| Line 2: | Line 2: | ||
If <math>\phi</math> is a homomorphism from ''G'' to ''H'' and <math>\sigma</math> is a homomorphism from ''H'' to ''K'', show that <math>\sigma\phi</math> is a homomorphism from ''G'' to ''K''. | If <math>\phi</math> is a homomorphism from ''G'' to ''H'' and <math>\sigma</math> is a homomorphism from ''H'' to ''K'', show that <math>\sigma\phi</math> is a homomorphism from ''G'' to ''K''. | ||
| + | |||
| + | ---- | ||
| + | |||
| + | |||
| + | for this problem you need to show that <math>\sigma</math>(<math>\phi</math>(a*b))=<math>\sigma</math>(<math>\phi</math>(a))*<math>\sigma</math>(<math>\phi</math>(b)). | ||
| + | |||
| + | The left hand side is in G and the right is in K. To prove this you must show a middle step that is created in H by applying the <math>\phi</math> to a*b to get <math>\sigma</math>(<math>\phi</math>(a*b))=<math>\sigma</math>(<math>\phi</math>(a)*<math>\phi</math>(b)). | ||
| + | |||
| + | Similarly, the same property can be applied to get to the final step. | ||
| + | --[[User:Podarcze|Podarcze]] 12:12, 18 February 2009 (UTC) | ||
Latest revision as of 07:12, 18 February 2009
If $ \phi $ is a homomorphism from G to H and $ \sigma $ is a homomorphism from H to K, show that $ \sigma\phi $ is a homomorphism from G to K.
for this problem you need to show that $ \sigma $($ \phi $(a*b))=$ \sigma $($ \phi $(a))*$ \sigma $($ \phi $(b)).
The left hand side is in G and the right is in K. To prove this you must show a middle step that is created in H by applying the $ \phi $ to a*b to get $ \sigma $($ \phi $(a*b))=$ \sigma $($ \phi $(a)*$ \phi $(b)).
Similarly, the same property can be applied to get to the final step. --Podarcze 12:12, 18 February 2009 (UTC)
