Question

Help.. Abstract Algebra problem

Answer 1

If A is isomorphic to B and C is isomorphic to D then show that
\[A \times C \approx B \times D\]

Answer 2

@freckles

Answer 3

@ganeshie8

Answer 4

try @zzr0ck3r

Answer 5

@IrishBoy123

Answer 6

or @Michele_Laino ??

Answer 7

@Michele_Laino

Answer 8

@freckles

Answer 9

Ok I might not be any good at this...
Just looking up definitions right now.
We are given:
\[f:A \rightarrow B \text{ is one to one and } f(x_1 \cdot x_2)=f(x_1) \cdot f(x_2) \text{ for all } x_1,x_2 \in A\]
and
\[g:C \rightarrow D\text{ is one to one and } g(y_1 \cdot y_2)=g(y_1) \cdot g(y_2) \text{ for all } y_1,y_2 \in C\]
We need to show:
\[h:A \times C \rightarrow B \times D \text{ is one to one and } h((x_1,y_1) \cdot h(x_2,y_2)))=h(x_1,y_1) \cdot h(x_2,y_2) \\ \text{ for all } (x_1,y_1)\]

Answer 10

oops for all (x1,y1) and (x2,y2) in AxC

Answer 11

*\[h:A \times C \rightarrow B \times D \text{ is one to one and } \\ h((x_1,y_1) \cdot h(x_2,y_2)))=h(x_1,y_1) \cdot h(x_2,y_2) \\ \text{ for all } (x_1,y_1) \text{ and } (x_2,y_2) \in h\]

Answer 12

ono-one and onto as well

Answer 13

\[h:A \times C \rightarrow B \times D \text{ means } h(x_1,y_1)=(f(x_1),g(y_1)) \text{ for all } (x_1,y_1) \in A \times C\]

Answer 14

for one to one
we want to see if \[h(x_1,y_1)=h(x_2,y_2) \implies (x_1,y_1)=(x_2,y_2)\]

Answer 15

correct

Answer 16

\[h(x_1,y_1)=h(x_2,y_2) \\ (f(x_1),g(y_1))=(f(x_2),g(y_2)) \\ \implies f(x_1)=f(x_2) \text{ and } g(y_1)=g(y_2)\]
guess what we already know f and g are one to one
this was givne

Answer 17

given*

Answer 18

ok.. now the onto part? The group preservation part is very simple

Answer 19

so we need to show this part...
\[h((x_1,y_1) \cdot (x_2,y_2))=h(x_1,y_1) \cdot h(x_2,y_2) \text{ where} (x_1,y_1) \text{ and }(x_2,y_2) \in A \times C\]

Do you know what the dot means between (x1,y1) and (x2,y2) ?

Answer 20

oh wait nevermind

Answer 21

that part i did already

Answer 22

\[h((x_1,y_1) \cdot (x_2,y_2)) \\ h((x_1 x_2, y_1 y_2)) \\ \text{ remember we were given } f(x_1 x_2)=f(x_1 ) \cdot f(x_2) \\ \text{ and } g(y_1 y_2)=g(x_1) \cdot g(x_2)\]

Answer 23

its the direct product of the elements

Answer 24

yeah correct

Answer 25

\[\text{ and that } h(a,b)=(f(a),g(b))\]


so we have...
\[h((x_1x_2,y_1y_2)=(f(x_1 x_2),g(y_1y_2))=(f(x_1) \cdot f(x_2)), g(y_1) \cdot g(y_2))\]

Answer 26

so we have:
\[h((x_1,y_1) \cdot (x_2,y_2))=(f(x_1) \cdot f(x_2),g(y_1) \cdot g(y_2))\]

and we want to show
this is is also equal to
\[h(x_1,y_1) \cdot h(x_2,y_2)\]

Answer 27

\[h(x_1,y_2) \cdot h(x_2, y_2)=(f(x_1),g(y_2)) \cdot (f(x_2), g(y_2))\]
we are basically done

Answer 28

I made a type-o that first y2 was suppose to be y1

Answer 29

its ok.. but i need the onto part

Answer 30

\[h(x_1,y_1) \cdot h(x_2, y_2)=(f(x_1),g(y_1)) \cdot (f(x_2), g(y_2))\]

Answer 31

ok I completely missed that little word onto

\[\text{ so we need to show for all } (b,d) \in B \times D \text{ there exist } (a,c) \in A \times C \\ \text{ such that } h(a,c)=(b,d)\]

we are given the following:
\[\text{ for all } b \in B \text{ there exist } a \in A \text{ such that } f(a)=b \\ \text{ for all } d \in D \text{ there exist } c \in D \text{ such that }g(c)=d \\ \text{ recall } h(a,c)=(f(a),g(c)) \]

Answer 32

but f(a)=b and g(c)=d
so that part wasn't hard at all

Answer 33

there exist c in C*

Answer 34

thanks.... :) thats a brilliant solution...:)

Answer 35

thanks kindly

Answer 36

you need to show you "map" is map. It is generally not obvious that a relation is a function. i.e. you need to show that your relation is defined everywhere and well defined.