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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 \(\mathbb Z_2\text{.}\)
