First isomorphism theorem

Algebra Level 2

Let ${\mathbb C}^*$ be the group of nonzero complex numbers under multiplication. Consider the homomorphism $f \colon {\mathbb C}^* \to {\mathbb C}^*$ given by $f(z) = z^2.$ Then the first isomorphism theorem says that ${\mathbb C}^*/\text{ker}(f) \simeq \text{im}(f).$

Now $\text{ker}(f) = \{ \pm 1\},$ and $\text{im}(f) = {\mathbb C}^*$ (that is, $f$ is onto), so the conclusion is that ${\mathbb C}^*/\{\pm 1\} \simeq {\mathbb C}^*,$ i.e. ${\mathbb C}^*$ is isomorphic to a nontrivial quotient of itself.

What is wrong with this argument?

f

is not onto Nothing is wrong; the statement is true

f

is not a homomorphism The kernel of

f

is not

\{\pm 1\}

1 solution

Patrick Corn
Aug 2, 2016

Nothing is wrong; the statement is true.

Not much to say: the argument is airtight, if I do say so myself! Note that the same argument shows that ${\mathbb C}^* / \mu_n \simeq {\mathbb C}^*,$ where $\mu_n$ is the set of $n$ th roots of unity ; just replace $z^2$ by $z^n$ in the definition of the homomorphism.

It may seem strange that a group can be isomorphic to a nontrivial quotient of itself, but it is completely possible.

Note that the same argument does not work if ${\mathbb C}$ is replaced by $\mathbb R,$ because the corresponding homomorphism will not be onto: not every real number has a real square root. (Exercise: What does the First Isomorphism Theorem say in this case?)

First isomorphism theorem

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