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Is there an intuitive explanation for the morphism theorem from introductory abstract algebra?

First Morphism Theorem: Let $K$ be the kernel of the group morphism $f: G \to H$. Then $G/K$ is isomorphic to $\text{Im } f$, and the isomorphism $\psi : G/K \to \text{Im } f$ is defined by $\psi(Kg) = f(g)$.

I follow the steps of the proof of the theorem, but find the theorem very nonintuitive to conceptualize.

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Do you find the concept of quotient group intuitive, or are you looking for an intuition of that concept too ? –  Joel Cohen Mar 19 '13 at 22:46
    
I understand the definition of the quotient group, but I don't really understand it intuitively, other than the trivial case of Z/nZ. If this would help me understand the morphism theorem, I would appreciate it. –  WeierstrassSauce Mar 19 '13 at 23:10
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4 Answers

up vote 7 down vote accepted

Quotients bunch up elements elements that are the same in some way, up into one thing. Quotienting $\mathbb{Z}$ by the subgroup $3\mathbb{Z}$ puts all things that have the same remainder after dividing by 3, into one set. So -5, 1 and 7 are all different elements in $\mathbb{Z}$, and the coset $$\{-5,-2,1,4,7\cdots\}=1+3\mathbb{Z}$$ puts everything with remainder 1 after division by 3, into one set.

Now consider that morphism $f:G\to H.$ Let's try to get an isomorphism from it. We want to get a bijective function. We can make it surjective imply by restricting the image. So $f^*:G\to \text{Im}f$ given by $f^*(g)=f(g)$ is a surjective homomorphism.

Okay, but how do we make it injective? Many things are getting mapped to the same image, because for any element $k$ of the kernel, $f(kg) = f(g).$ In fact, the pre-image of $f(g)$ is $Kg.$ To make it injective, we could put everything that gets mapped to $f(g)$ together so that by construction, only one thing gets mapped to each element of the image (i.e being injective). This means precisely considering the quotient $G/K.$ So now if we have $\phi:G/K \to \text{im} f$ defined by $ \phi(Kg) = f(g)$ then it is certainly surjective, a morphism and injective as well because we clumped all the things that used to get sent to $f(g)$ all up into one element.

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+1, very nice explanation. –  Martin Brandenburg Mar 19 '13 at 23:45
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This is very nice and concise, thanks! –  WeierstrassSauce Mar 21 '13 at 2:57
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I hope that I can convince you that quotient groups arise naturally and that the isomorphism theorem is a very basic and easy observation.

Let $p: X \to Q$ be a map of sets. Really, just sets. Let $f : X \to Y$ be another map. When can we extend $f$ to a map $\tilde{f} : Q \to Y$, i.e. a map satisfying $\tilde{f} \circ p = f$? In other words, we want to have $\tilde{f}(p(x))=f(x)$. The uniqueness is clear when $p$ is surjective, because then every element of $Q$ has the form $p(x)$ for some $x \in X$, but we already know $\tilde{f}$ on these elements. Conversely, we have to check when $\tilde{f}(p(x)) := f(x)$ declares a well-defined function, i.e. that $p(x)=p(x')$ implies $f(x)=f(x')$. In other words, $\ker(p) \subseteq \ker(f)$, where $\ker(f)$ is the equivalence relation on $X$ which is defined by $(x,x') \in \ker(f) \Leftrightarrow f(x)=f(x')$. These trivial observations show:

If $p : X \to Q$ is a surjective map, then a map $f : X \to Y$ extends uniquely to a map $\tilde{f} : Q \to Y$ if and only if $\ker(p) \subseteq \ker(f)$. We have $\mathrm{im}(f)=\mathrm{im}(\tilde{f})$ and $\tilde{f}$ is injective if and only if $\ker(p)=\ker(f)$.

Besides, if $\sim$ is an equivalence relation on $X$, then the projection $p : X \to X/\sim$ on the set of equivalence classes has by definition $\sim$ as kernel. Now, the same holds for all algebraic structures. For example, we have:

If $p : G \to Q$ is a surjective homomorphism of groups, then a homomorphism $f : G \to H$ extends uniquely to a homomorphism $Q \to H$ if and only if $\ker(p) \subseteq \ker(f)$.

The proof uses the set case above and the observation that $\tilde{f}$ is a homomorphism iff $f$ is a homomorphism, since $p$ is surjective.

In the case of groups, we have $(g,g') \in \ker(f) \Leftrightarrow f(g)=f(g') \Leftrightarrow f(g g'^{-1})=1 \Leftrightarrow g g'^{-1} \in \mathrm{Ker}(f)$, where the latter is the kernel in the usual group theoretic sense. From this we see that $\ker(p) \subseteq \ker(f)$ holds iff $\mathrm{Ker}(p) \subseteq \mathrm{Ker}(f)$. Every normal subgroup is the kernel of a surjective homomorphism of groups - this is what the construction of quotient groups is all about. Thus:

If $G$ is a group and $N$ is a normal subgroup of $G$, then there is a surjective group homomorphism $p : G \to G/N$ with $\mathrm{Ker}(p)=N$, with the following property: A homomorphism $f : G \to H$ extends uniquely to a homomorphism $\tilde{f} : G/N \to H$ if and only if $N \subseteq \mathrm{Ker}(f)$. We have $\mathrm{im}(\tilde{f})=\mathrm{im}(f)$ and $\tilde{f}$ is injective if and only if $N = \mathrm{Ker}(f)$.

This is one of the first universal properties one encounters when studying mathematics. Universal properties help to understand the real nature of mathematical objects. In this case, it is not important at all (and in fact, often just misleading) that $G/N$ consists of cosets. The universal property above is really what is used and what is important when working with quotient groups. It expresses exactly the idea that $G/N$ is the "smallest extension" of $G$ in which the elements are "killed".

For example, let us start with the abelian group $\mathbb{Z}$. Imagine all the integers lined up. Now we want to "kill" some positive integer $n$, i.e. we want to force $n=0$. This means that we make the line to a circle, since after $0,1,\dotsc,n-1$ we get again the same $n=0,n+1=1,\dotsc$ etc. This is formalized in the construction of the quotient group $\mathbb{Z}/n\mathbb{Z}$.

As a special case of the universal property above we get:

If $f : G \to H$ is a homomorphism of groups, there is a unique isomorphism $\tilde{f} : G/\mathrm{Ker}(f) \cong H$ satisfying $\tilde{f} \circ p = f$, where $p : G \to G/\mathrm{Ker}(f)$ is the projection.

This is the first isomorphism theorem. As you can see, the "proof" is really just a bunch of trivial observations. But the significance of this theorem cannot be overestimated. As already indicated, it holds for arbitrary algebraic structures. This theorem is used on almost every page of a modern book on algebra. Therefore I won't try to list all the applications.

Let me just mention two special cases:

  1. Let $n \geq 2$. The signum of a permutation is a homomorphism $\mathrm{sgn} : S_n \to \{\pm 1\}$. It is surjective (the transposition $(1 2)$ has signum $-1$) and the kernel is, by definition, the alternating group $A_n$. Thus, $S_n / A_n \cong \{\pm 1\}$.

  2. Intuitively, $\mathbb{R}/\mathbb{Z}$ should be a circle (since here the integers are identified and therefore the real line repeats itsself again and again, resulting in a circle). In fact, there is a homomorphism $(\mathbb{R},+) \to (\mathbb{C}^*,*)$, $t \mapsto \exp(2\pi i t)$. It has image $S^1 = \{z \in \mathbb{C} : |z|=1\}$ and the kernel is $\mathbb{Z}$. Thus, we have $\mathbb{R}/\mathbb{Z} \cong S^1$.

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Maybe look at some simple examples.

Consider the groups $G = (\mathbb{Z},+)$ and $H = (\mathbb{C}^*, \times)$. Define $f : G \to H$ by $f(n) = (-1)^n$ for $n \in \mathbb{Z}$. You can check that $f$ is a group morphism. We can see that $\textrm{Im}(f) = \{-1,1\}$ and $\ker(f) = 2\mathbb{Z}$. In that case, the theorem states that

  • The value of $f(k) = (-1)^k$ only depends the class of $k$ modulo $\ker(f) = 2\mathbb{Z}$ (i.e. the parity of $k$) : it's $-1$ if $k$ is odd, and $1$ if $k$ is even.
  • The previous point allows you to define $\psi : \mathbb{Z}/2\mathbb{Z} \to \{-1,1\}$ by $\psi(k + 2 \mathbb{Z}) = (-1)^k$ and this is a group isomorphism.

For another (similar) example, fix $n \in \mathbb{N}^*$ and define $f : G \to H$ by $f(k) = e^{\frac{2ik\pi}{n}}$. Denote $\mathbb{U}_n = \left\{1, e^{\frac{2i\pi}{n}}, e^{\frac{2i2\pi}{n}}, \ldots,e^{\frac{2i(n-1)\pi}{n}}\right\} = \{z \in \mathbb{C}, z^n = 1\}$. Then $\textrm{Im}(f) = \mathbb{U}_n$, $\ker(f) = n \mathbb{Z}$ and the theorem yields an isomorphism $\psi : \mathbb{Z}/n\mathbb{Z} \to \mathbb{U}_n$ defined by $\psi(k+n\mathbb{Z}) = e^{\frac{2ik\pi}{n}}$ (as in the case $n = 2$, the value $e^{\frac{2ik\pi}{n}}$ only depends on the class of $k$ modulo $n$, so $\psi$ is well defined).

In a similar fashion, consider $G = (\mathbb{R},+)$, $H = (\mathbb{C}^*, \times)$ and $f: x \mapsto e^{i x}$. We have $\text{Im}(f) = \{z \in \mathbb{C}, |z| = 1\}$ and $\ker(f) = 2\pi\mathbb{Z}$. In that case the theorem gives you an isomorphism $\psi : \mathbb{R}/2\pi\mathbb{Z} \to \{z \in \mathbb{C}, |z| = 1\}$ given by $\psi(x + 2\pi\mathbb{Z}) = e^{i x}$ (once again this is well defined because $e^{ix}$ only depends on "$x$ modulo $2\pi$"). As a matter of fact, the reciprocal of $\psi$ in that case is what we usually call the argument (the argument of a complex number is only defined "up to a multiple of $2\pi$", which essentially means it really is an element of $\mathbb{R}/2\pi\mathbb{Z}$).

For yet another example, consider $G = H = (\mathbb{C}^*, \times)$. Fix $n \in \mathbb{Z}$ with $n \ne 0$ and define $f : G \to H$ by $f(z) = z^n$. Then $f$ is a morphism and $\textrm{Im}(f) = \mathbb{C}^*$ (the equation $x^n = y$ always has a solution because $\mathbb{C}$ is algebraically closed) and $\ker(f) = \mathbb{U}_n$. The theorem gives you an isomorphism $\psi : \mathbb{C}^*/\mathbb{U}_n \to \mathbb{C}^*$ defined by $\psi(z\mathbb{U}_n) = z^n$. You could think of of $\psi^{-1}(z)$ as $\sqrt[n]{z}$ but because there are $n$ possible choices for an $n^{\textrm{th}}$ root of $z$ (all of which only differ by an element of $\mathbb{U}_n$), it is not a well defined number, but a class modulo $\mathbb{U}_n$.

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Here's an attempt to explain by example. Take the real line $\Bbb{R}$, and fold it around the unit circle like this: The interval $[0, 2\pi]$ folds once counterclockwise around the circle. The interval $[2\pi, 4\pi]$ folds once again, overlapping it. Continue this infinitely many times with the remaining intervals $[4\pi, 6\pi], ...$ and also $[-2\pi, 0], ...$.

The function that tells us where a point in $\Bbb{R}$ goes to on the unit circle is a homomorphism of the group $\Bbb{R}$ with the operation $+$ to the group $S^1$ (the unit circle) with the rotation operation. (This is since adding $\theta$ in $\Bbb{R}$ corresponds to rotating by $\theta$ radians on the circle.)

(You may also think of the unit circle as the set of complex numbers with modulus $1$. This is a group with the complex number product as the group operation.)

Let's see what the theorem says about this homomorphism:

Let $K$ be the kernel of the group morphism...

Okay, so $K$ is a subgroup of $\{\Bbb{R}, +\}$ which consists of all real numbers that go to the point on the circle with angle $0$. By our construction, $K = 2\pi \Bbb{Z}$, meaning it's all integer multiples of $2\pi$. (You may check that this is a subgroup.)

Then $G/K$ is isomorphic to $\text{Im } f$...

The image $\text{Im }f$ is easy - this is simply the unit circle. So, the theorem says the unit circle is isomorphic to the quotient group $\Bbb{R}/2\pi\Bbb{Z}$. Time to remember what the quotient group is.

The quotient group $\Bbb{R}/2\pi\Bbb{Z}$ consists of elements of the form $r + 2\pi\Bbb{Z}$, where $r$ is a real number. You may notice that $r$, $r+2\pi$, $r+4\pi$, and so on, all go to the same element of the group. So adding or subtracting $2\pi$ to members of the quotient group is like rotating the circle by $2\pi$. Indeed - this sounds isomorphic.

So, our morphism divides the information about real numbers to two parts:

  1. Which interval they're in (out of $[0, 2\pi], [2\pi,4\pi]$ and so on...), and
  2. where they are in the interval.

Notice that the kernel $K$ corresponds exactly to the first piece of information and the quotient group $G/K$ corresponds to the second piece of information. So, the theorem tells us that if we "delete" from our group the information in the morphism's kernel, the remaining information is exactly the same as the image of the morphism.

...and the isomorphism $\psi:G/K\to\text{Im }f$ is defined by $\psi(Kg)=f(g)$.

In our case, this is $\psi(r+2\pi\Bbb{Z})=f(r)$. Meaning you can take a the position of a real number in one of the intervals (forgetting which exact interval it was in), and use that to find the position on the circle that the point goes to - and this is an isomorphic map.

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