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Problem 90 (A homomorphism is an $n$-to-one map)
Let $f:G\to H$ be a homomorphism.
- Suppose that $\ker f$ has order $n$. If we know that $f(g)=h$ for some $g\in G$, prove that there are exactly $n$ different elements in $G$ that map to $h$, so we know that the preimage of $h$, namely $f^{-1}(h)$, is a set with $n$ elements. This shows that $f$ is an $n$-to-one map.
- Prove that $f$ is injective if and only if the kernel of $f$ is trivial, i.e. $\ker f = \{e\}$.
- Prove that if $f$ is surjective and the kernel of $f$ is trivial, then $f$ is an isomorphism.
Here are some comment's about these problems.
- This should follow without any additional work from a previous problem. Just figure out which problem.
- This is why in linear algebra we know that $A\vec x = b$ has a single solution if and only if the only solution to $A\vec x=\vec 0$ is $\vec x= 0$, which means the columns of $A$ are linearly independent. Finding the kernel, or null space, is crucial in linear algebra.
- This is why in linear algebra we know that when the kernel of a square matrix is trivial, then the matrix is invertible.
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