First Isomorphism Theorem

**mathbeginner** · November 1st 2010, 10:28 AM

Let S,T be subspaces of the vector spaces V ,U (respectively), and let

$T : V × U -> (V/S)× (U/T)$ be the linear map defined by $T(x, y) = (x +S, y + T).$ Find the kernel of T and prove that

$(V× U)/(S× T) (~ =) (V/S) × (W/T).$

Plx help...

**roninpro** · November 1st 2010, 11:56 AM

First, do you mean to define $T(x, y) = (x +S, y + T)$ ?

I think that your first step is to determine the kernel of $T$ . If $T(x,y)=(0 +S, 0 +T)$ , what can be said about $x$ and $y$ ?

**mathbeginner** · November 1st 2010, 12:31 PM

Originally Posted by roninpro

First, do you mean to define $T(x, y) = (x +S, y + T)$ ?

I think that your first step is to determine the kernel of $T$ . If $T(x,y)=(0 +S, 0 +T)$ , what can be said about $x$ and $y$ ?

is it Ker T= T(0,0)=(0+S, 0+T) =(S,T)
therefor Ker T = (S,T)?

but I still don't know how to prove the following....

**Drexel28** · November 1st 2010, 01:27 PM

Originally Posted by mathbeginner

is it Ker T= T(0,0)=(0+S, 0+T) =(S,T)
therefor Ker T = (S,T)?

but I still don't know how to prove the following....

You have that $F:V\boxplus U\to (V/S)\boxplus (U/T)<img src=$ x,y)\mapsto \left(x+S,y+T)" alt="F:V\boxplus U\to (V/S)\boxplus (U/T)

x,y)\mapsto \left(x+S,y+T)" /> (I use $\boxplus$ to differentiate between external and internal direct sum), this is evidently a surjective linear homomorphism. Then, we are trying to look for what values $(x,y)$ is $T(x,y)=...$ ...equals what? We're trying to look for the origin in the vector spaces $V/S$ and $U/T$ but these are precisely $S$ and $T$ respectively. So, for what $x$ is $x+S=S$ and for what $y$ is $y+T=T$ ? It is fairly clear that these statements are true if and only if $x\in S$ and $y\in T$ . Thus, a quick proof shows that $\ker F=S\boxplus T$ and so by the F.I.T. $\left(V\boxplus U\right)/\left(S\boxplus T\right)\cong \left(V/S\right)\boxplus\left(U/T\right)$

**mathbeginner** · November 1st 2010, 04:20 PM

I don't reli get why Ker F = S X T and how to prove that???

**Drexel28** · November 1st 2010, 04:26 PM

Originally Posted by mathbeginner

I don't reli get why Ker F = S X T and how to prove that???

Come on friend. Review your definitions. You're trying to show that $\ker F=S\boxplus T$ which is equivalent to showing $F(x,y)=(S,T)\Leftrightarrow (x,y)\in S\boxplus T$ . I give the solution below "hidden". Try your damnedest, and only look at it as a last result.
[spoil]Clearly if $(x,y)\in S\boxplus T$ then $F(x,y)=(x+S,y+T)=(S,T)$ since $x+S=\left\{x+s:s\in S\right\}=S$ since $S\leqslant V$ and similarly $y+T=T$ . Conversely, suppose that $latex F(x,y)=(S,T)[/tex] then $x+S=S$ . But, note that $x\in x+S$ since $0\in S$ and so $x\in S$ . Similarly $y+T=T\implies y\in T$ and so $F(x,y)=(S,T)\implies (x,y)\in (S,T)$ . Combining these we get $F(x,y)=(S,T)\Leftrightarrow (x,y)\in S\boxplus T$ [/spoil]

EDIT: I can't figure out how to hide it...so work on the honor code haha

**TheCoffeeMachine** · November 1st 2010, 04:53 PM

Originally Posted by Drexel28

EDIT: I can't figure out how to hide it...so work on the honor code haha

[Spoiler] [/Spoiler] You have been gone for so long that you forgot the tags!

**Drexel28** · November 1st 2010, 05:01 PM

Originally Posted by TheCoffeeMachine

[Spoiler] [/Spoiler] You have been gone for so long that you forgot the tags!

Haha, I guess so!

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