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I've been studying homomorphisms and have got to the first isomorphism theorem. The proof generally makes sense but I'm not able to see why the theorem is intuitively true, nor how it could be useful.

I was hoping someone could go through the proof and explain the 'intuitiveness' of each step and why it should 'feel' true.

So the theorem and the structure of the proof I've is as follows:

Let $\phi: G_1 \rightarrow G_2$ be a group homomorphism Let $\ker(\phi)$ be the set $\left\{x \in G_1 | \phi(x) = 0_{G_2}\right\}$ Then $\left\{ y \in G_2 | \exists g \in G_1, \phi(g) = y \right\}$ = im$(\phi)$ is isomorphic to the set of cosets of $\ker(\phi)$ in $G$.

Proof: Show that the map $\psi: G / K \rightarrow im(\phi)$ defined by $\psi(g+K) = \phi(g)$ is well defined. Prove that this map is surjective Prove that this map is injective by showing the kernel is simply the identity.

Note that in the module I'm taking every group is assumed to be abelian thus we don't study normal subgroups or distinguish between left and right cosets.

A few other questions:

1) What 'is' $G / K$? I can recite the definition but it feels very arbitrary to me. I get the feeling it's kind of like splitting G into partitions of elements that don't map to zero?

2) Why can we map sets of elements (cosets) onto specific elements of $G_2?$ My understanding of isomorphism is that they essentially represent the same thing, but one side is a set and the other is an element?

3) Is there an alternative map I can use from $im(\phi) \rightarrow G/K$ to prove the isomorphism?

Thank you

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  • $\begingroup$ $G/K = \{Kg|g\in G\}$ i.e. the set of all (right (and left in your case)) cosets of G. $\endgroup$ – fosho Jan 3 '16 at 20:11
  • $\begingroup$ Related: Need isomorphism theorem intuition. $\endgroup$ – Andrew D. Hwang Jan 3 '16 at 21:04
  • $\begingroup$ Hi I really liked your analogy in that post, would you be able to expand upon it? $\endgroup$ – Kevin Jan 4 '16 at 13:23
  • $\begingroup$ @Kevin: Assuming you meant the dark sunglasses analogy, this is a general property of equivalence relations. (When you look at a set $G$ through an equivalence relation $R$, you see $G/R$, the set of equivalence classes; there's not really more to say.) The answer on the linked post explains why level sets of a homomorphism $\varphi$ are precisely the left cosets of the kernel $K = \ker(\varphi)$ in the simplest manner I know. Did you have anything specific in mind by "expand"? $\endgroup$ – Andrew D. Hwang Jan 4 '16 at 22:48
  • $\begingroup$ So is the idea that we think of $\phi$ as our sunglasses which is applied to $G_1$, now we are looking at the image of $G_2$. But all of the elements can only be seen in terms of cosets, thus we relable each coset (by constructing an isomorphism) so we can properly see $G_2.$ $\endgroup$ – Kevin Jan 5 '16 at 19:04
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Let's say we have a homomorphism $\phi : G \to H$, where $K = \ker \phi$, i.e. $K = \{ g \in G : \phi(g) = 0_H\}$. The intuition of the quotient group $G/K$ is that you "fold and glue" the domain so that all of the points with the same image are glued together. Because all of the elements with the same image are glued together - and so become the same thing - our homomorphism becomes injective, i.e. becomes an isomorphism.

To (try to) answer your questions:

1) The object $G/K$ is given by the elements of $G$ modulo the elements of $K$. For example, for any $g$ in $G$ we get an equivalence class of elements of $G$, namely $[g] = g+K = \{g+k : k \in K\}$. The whole point here is that all of $[g]$ get sent to the same thing by $\phi$.

2) The element $[g] \in G/K$ is given by all elements of $G$ of the form $g+k$ where $k \in K$. Since the mapping $\phi : G \to H$ is a homomorphism, we have $\phi(g+ k) = \phi(g)+\phi(k) = \phi(g) + 0_H=\phi(g).$ For any $[g] \in G/K$ the mapping $\varphi : G/K \to H$ is defined by $\varphi([g]) = \phi(g)$. Remember that "isomorphism" means "one/same structure". It doesn't mean that the things in the domain and range are the same sort of things. It means that, as groups, the domain and range have exactly the same group structure.

3) Instead of "$im(G) \to G/K$", why not use $\varphi : G/K \to \phi(G)$ which comes from $\phi : G \to H$ be putting $\varphi([g])=\phi(g)$ as your isomorphism, where $\phi([g]) = \phi(g)$. It's easy to prove from the definition $[g]=g+K$ that $$\varphi([g_1]+[g_2])=\varphi([g_1+g_2])=\phi(g_1+g_2)=\phi(g_1)+\phi(g_2)=\varphi([g_1])+\varphi([g_2])$$ Also, notice that, $\ker \varphi = \{ [g] \in G/K : \varphi([g]) = 0_H \}$. Moreover, $$\varphi([g])=0_H \iff \phi(g+K) = 0_H \iff \phi(g) = 0_H \iff g \in K$$ It follows that $\ker \varphi = [k]$, where $k \in K$, and so $\varphi$ is injective.

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  • $\begingroup$ Thank you. With regards to the first part about 'fold and glue' the domain. I can see that we are seperating the points that map to zero, but how can we tell that the rest of the map, or the different equivalence classes remain injective? I've tired doing $\phi(g+K) = \phi(h+K) \Rightarrow (g-h) \in K$. So after we have 'split' G into partitions of elements that map to distinct points, each of those distinct points will map to a part of the image of $\phi$ thus a bijection seems intuitive. Am I anywhere close? $\endgroup$ – Kevin Jan 4 '16 at 13:21
  • $\begingroup$ That's right! Well, you get a bijection on to the image $\varphi(G/K)$. $$\varphi([g_1]) = \varphi([g_2]) \iff \phi(g_1+K) = \phi(g_2+K) \iff \phi(g_1)+\phi(K) = \phi(g_2)+\phi(K) \\ \iff \phi(g_1) + 0_H = \phi(g_2) + 0_H \iff \phi(g_1) = \phi(g_2)$$ $\endgroup$ – Fly by Night Jan 5 '16 at 20:58

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