# Proof of The Third Isomorphism Theorem

Mathematics Asked by Abhijeet Vats on August 12, 2020

Here’s what I’m trying to prove right now:

Let $$V$$ be a vector space over $$mathbb{F}$$. Let $$M$$ be a linear subspace of $$V$$ and $$N$$ be a linear subspace of $$M$$. Prove that the mapping $$x+N mapsto x+M$$ between the quotient spaces $$V/N to V/M$$ is linear with kernel $$M/N$$. Then, deduce that:

$$frac{V/N}{M/N} cong V/M$$

Proof Attempt:

We have to show that the given relation $$T: V/N to V/M$$ is a linear mapping. We define:

$$forall x in V: T(x+N) = x+M$$

This is totally-defined. To show well-definedness, let $$x+N = y+N$$ where $$x,y in V$$. Then, $$x-y in N$$. So, $$x-y in M$$. Hence:

$$x + M = y+M$$

$$iff T(x+N) = T(y+N)$$

To prove that it is linear, we need to show additivity and homogeneity.

Let $$u,v in V/N$$. Then, $$u = x+N$$ and $$v = y+N$$ for some $$x,y in V$$. Then:

$$T(u+v) = T((x+N)+(y+N)) = T((x+y)+N) = (x+y)+M = (x+M)+(y+M) = T(u) + T(v)$$

1. Proof of Homogeneity

Let $$alpha in mathbb{F}$$ and $$u in V/N$$. Then, $$u = x+N$$ for some $$x in V$$. So:

$$T(alpha u) = T(alpha(x+N)) = T(alpha x +N) = alpha x + M = alpha (x+M) = alpha T(u)$$

This proves homogeneity. Hence, $$T$$ is a linear map. To show that the kernel of $$T$$ is $$M/N$$, we have:

$$T(x+N) = x+M = theta_V+M$$

$$iff x in M$$

$$iff x+N in M/N$$

$$iff ker(T) = M/N$$

Now, we notice that $$T$$ is surjective. By the first isomorphism theorem, it follows that:

$$frac{V/N}{M/N} cong V/M$$

That proves the desired result.

Does the proof above work? If it doesn’t, why? How can I fix it?

I would change the part concerning $$ker T$$, writing: begin{align} x + N in ker T & iff T(x+N) = 0 in V/M \ & iff x+M = 0 in V/M \ & iff x in M \ & iff x + N in M/N end{align} which means $$ker T = M/N$$.

Correct answer by Rodrigo Dias on August 12, 2020

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