Theorem 5.
Let \(G\) and \(H\) be groups and \(\psi: G \to H\) a surjective homomorphism with kernel \(K\text{.}\) Then
\begin{equation*}
G/K \cong H\text{.}
\end{equation*}
That is, the factor group \(G/K\) is isomorphic to \(H\text{.}\)
Section Week 13 (11/17-11/21)
Handout Monday 11/17
More investigation of homomorphisms today. After an introduction about what a homomorphic image is, work on the Homomorphic Images activity.
If there is time after discussing the activity, consider what happens with \(D_4\text{.}\) That is, which groups are homomorphic images of \(D_4\text{?}\)
Handout Wednesday 11/19
We have been looking at homomorphisms. A natural question:
We have also been considering quotient groups. Note that when \(H\) is a normal subgroup of \(G\text{,}\) we can define \(G/H = \{aH ~:~a \in G\}\) and this is a group under the operation of coset multiplication (addition) given by
\begin{equation*}
aH bH = abH.
\end{equation*}
This looks a bit like the homomorphism property! Is \(G/H\) a homomorphic image of \(G\text{?}\)
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Yes: we can define the “obvious” homomorphism \(f:G \to G/H\) by \(f(a) = aH\text{.}\) This is called the canonical homomorphism. Check that this really is a homomorphism.
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What is the kernel? It is just \(H\text{!}\)
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This is great: every homomorphism has a kernel that is a normal subgroup. This says every normal subgroup is the kernel of some homomorphism.
We know that for any group \(G\) with normal subgroup \(K\text{,}\) the quotient group \(G/K\) is a homomorphic image of \(G\text{.}\) Are there any other homomorphic images of \(G\text{?}\) In general, if \(H\) is a homomorphic image of \(G\text{,}\) what relationship does \(H\) have to some quotient group.
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The Fundamental Homomorphism Theorem: if \(f:G \to H\) is a surjective homomorphism with kernel \(K\text{,}\) then \(G/K \cong H\text{.}\)
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In other words, every homomorphic image of a group \(G\) is isomorphic to a quotient group of \(G\) (by the kernel).
First, let’s prove a suspicion we have had about homomorphisms: if two elements are sent to the same output, then they must belong to the same coset of the kernel.
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That is, suppose \(\psi:G \to H\) is a homomorphism with kernel \(K\text{.}\) Suppose \(\psi(a) = \psi(b)\text{.}\)
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Then \(\psi(b\inv)\psi(a) = e\text{,}\) which means \(\psi(b\inv a) = e\text{.}\) That is, \(b\inv a \in K\text{.}\)
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We have seen this before: \(b\inv a = k_1\) so \(a = bk_1\) which says \(a \in bK\) and therefore \(aK = bK\text{.}\)
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The converse is also true: if \(aK = bK\) then \(b\inv a \in K\) so \(\psi(b\inv a) = e\text{.}\) This means \(\psi(a) = \psi(b)\text{.}\)
Proof.
We assume \(G\text{,}\) \(H\text{,}\) \(\psi\text{,}\) and \(K\) are as in the statement of the theorem. We must define a isomorphism \(\eta: G/K \to H\text{,}\) and prove it really is an isomorphism.
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Define \(\eta(xK) = \psi(x)\text{.}\) What???
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What does this say? Note \(\psi(x)\) is an element of \(H\text{;}\) the image of \(x\text{.}\) So we are saying that \(\eta\) sends the coset \(xK\) to the image of \(x\text{.}\)
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We should be careful. Is this even a function? Is it well defined? Suppose \(aK = bK\text{.}\) Is \(\eta(aK) = \eta(bK)\text{?}\)
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Then\begin{equation*} \eta(aK) = \psi(a) = \psi(bk) = \psi(b)\psi(k) = \psi(b) = \eta(bK)\text{.} \end{equation*}
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Is \(\eta\) a bijection?
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Suppose \(\eta(aK) = \eta(bK)\text{.}\) Then \(\psi(a) = \psi(b)\text{.}\) But if \(a\) and \(b\) are sent to the same element of \(H\text{,}\) then they must belong to the same coset of \(K\text{,}\) so \(aK = bK\text{.}\) Thus \(\eta\) is one-to-one.
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To see that \(\eta\) is onto, consider any \(h \in H\text{.}\) Since \(\psi\) is onto, there is some \(g \in G\) such that \(\psi(g) = h\text{.}\) So \(\eta(gK) = h\text{.}\)
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Finally, is \(\eta\) a homomorphism? Check the property!\begin{equation*} \eta(aK)\eta(bK) = \psi(a)\psi(b) = \psi(ab) = \eta(abK) = \eta(aKbK)\text{.} \end{equation*}
Handout Friday 11/21
Let’s prove that the isomorphism we defined last time is a bijection. This will be helpful for the last homework too, where you need to prove something is injective.
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Recall that we were given \(\varphi:G \to H\) with kernel \(K\text{.}\) We defined a new function \(\psi:G/K \to H\) by \(\psi(aK) = \varphi(a)\text{.}\) We already showed that this function is well defined and satisfies the homomorphism property. So it is at least a homomorphism. Is it an isomorphism?
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Suppose \(\psi(aK) = \psi(bK)\text{.}\) Then \(\varphi(a) = \varphi(b)\text{.}\) But if \(a\) and \(b\) are sent to the same element of \(H\text{,}\) then they must belong to the same coset of \(K\text{,}\) so \(aK = bK\text{.}\) Thus \(\psi\) is one-to-one.
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To see that \(\psi\) is onto, consider any \(h \in H\text{.}\) Since \(\varphi\) is onto, there is some \(g \in G\) such that \(\varphi(g) = h\text{.}\) So \(\psi(gK) = h\text{.}\)
Now some examples of how the FHT is used.
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One simple way the FHT helps us is to prove that particular groups are isomorphic.
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For example, there is a homomorphism from \(\mathbb Z_{12}\) to \(\mathbb Z_6\text{.}\) What is it? More importantly, what is its kernel? We have \(K = \{0, 6\} = \langle 6 \rangle\) so the FHT tells us that \(\mathbb Z_6 \cong \mathbb Z_{12}/\langle 6 \rangle\text{.}\)
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Perhaps a less obvious example is the following. Let \(\mathbb C_1\) be the set of all complex numbers \(a + bi\) such that \(a^2 + b^2 = 1\text{.}\) These are all the complex numbers on the unit circle in the complex plane.
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We can write each of these points in the form \(\cos(x) + i \sin(x)\) for real numbers \(x\text{.}\) We often use the shorthand \(\mathop{\mathrm{cis}}x\) for this.
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Using facts about trig functions, we can prove that \(\mathbb C_1\) is a group under multiplication.
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We can define a homomorphism from \(\mathbb R\) to \(\mathbb C_1\) by \(f(x) = \mathop{\mathrm{cis}}(2\pi x)\text{.}\) What is the kernel of \(f\text{?}\) It is simply \(\mathbb Z\text{.}\)
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The FHT tells us then that \(\mathbb C_1 \cong \mathbb R/\mathbb Z\text{.}\)
We can also use the FHT in a negative way - to prove that there cannot be a surjective homomorphism from a group \(G\) to a group \(H\text{.}\)
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No. If it were, then suppose the kernel was \(K\text{.}\) We would have \(\mathbb Z_3 \cong S_3/K\text{.}\) What can you say about the size of \(K\text{?}\)
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We must have \(|K| = 2\text{,}\) since \(S_3/K\) has order 3 and \(S_3\) has order 6. But \(K\) needs to be a normal subgroup of \(S_3\text{,}\) and there are no normal subgroups of \(S_3\) with size 2.
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More in general, we can ask, up to isomorphism, how many homomorphic images of \(S_3\) are there? In other words, how many different homomorphisms have \(S_3\) as their domain?
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Well, any homomorphism from \(S_3\) to some group \(H\) will have a kernel \(K\text{,}\) and then FHT promises us that \(H \cong S_3/K\text{.}\)
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This says that any two homomorphisms with the same kernel are really the same homomorphism!
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So how many possible kernels are there? Well, the kernel needs to be a normal subgroup, so in \(S_3\) the only possibilities are \(S_3\text{,}\) \(\{(1)\}\) and \(A_3 = \{(1), (123), (132)\}\text{.}\)
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So there are exactly three homomorphic images of \(S_3\text{:}\) \(S_3/S_3\text{,}\) \(S_3/\{(1)\}\text{,}\) and \(S_3/A_3\text{.}\) Every homomorphic image of \(S_3\) must be isomorphic to one of these - the first two are not particularly interesting - the trivial 1 element group and all of \(S_3\text{.}\) The third is isomorphic to \(\mathbb Z_2\text{.}\)
