Consider the determinant function Det:M$_{n}(\mathcal{F})$$\rightarrow\mathcal{F}$, where $\mathcal{F}$ is a field.

i) Explain how to restrict the domain and range of Det to

obtain a group homomorphism. State any important properties of the determinant

function that are used to prove that the resulting map is a homomorphism.

(Do not need to prove these properties).

ii) Is Det: M$_{3}(\mathcal{F})$$\rightarrow\mathcal{F}$ one-to-one, or onto , or both, or neither? Explain your

answer with examples.

For part i) of this question could the answer be that the domain could be the complex numbers but the range could be the complex numbers excluding 0, as if the output of the determinant was zero the matrix would not have an inverse so it would not be a group ?.

For part ii) proving that the function is one-to-one is routine but I do not know how we could prove or disprove that the determinant is onto? .

  • $\begingroup$ I have edited some of the details to help provide context and some of my thoughts on the problem. $\endgroup$
    – Massin
    Jul 2, 2014 at 4:55

1 Answer 1


I assume you know that $\det (AB) = \det A \det B$. So, the multiplicativity you need for a homomorphism is done. It remains to find the groups.

The largest multiplicative group in $F$ is $F^{\times}=F\setminus \{0\}$. The inverse image of $F^{\times}$ under $\det$ is the set of matrices that have non-zero determinant, which is the same as the set of invertible matrices. This set is a group, called the general linear group over $F$ and denoted by $GL_n(F)$.

Hence $\det$ induces a group homomorphism $GL_n(F) \to F^{\times}$.

This homomorphism is surjective but not injective. Look at diagonal matrices to prove both properties.

  • $\begingroup$ You have to restrict the domain of $\det$ because $\det(A+B) = \det(A) + \det(B)$ does not hold in general and so you cannot use the additive group of $F$. $\endgroup$
    – lhf
    Jul 2, 2014 at 14:10

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