1. Let $G=\mathrm{GL}_n(\mathbb R)$ and $H=\mathbb R^*$. Let $\phi : G=\mathrm{GL}_n(\mathbb R) \rightarrow \mathbb R^*$ be the map defined by $\phi(A)=\det(A)$. Show that $\phi$ is a group homomorphism.

  2. If $\phi : G \rightarrow H$ is a group homomorphism, then the set $\{g \in G : \phi(g)=e_H\}$ where $e_H$ is the identity element of $H$ is called the kernel of $\phi$. Show that the kernel of $\phi$ is a subgroup of $G$.


closed as too localized by tomasz, Jack Schmidt, Micah, Tom Oldfield, Amzoti May 20 '13 at 20:20

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  • $\begingroup$ What have you tried? (The first should be a well-known thing from linear algebra and the second is just applying the definition). $\endgroup$ – Tobias Kildetoft May 20 '13 at 19:25
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    $\begingroup$ To get the best possible answers, you should explain what your thoughts on the problem are so far. That way, people won't tell you things you already know, and they can write answers at an appropriate level; also, people tend to be more willing to help you if you show that you've tried the problem yourself. Many would consider your post rude because it is a command ("Show..."), not a request for help, so please consider rewriting it. $\endgroup$ – Zev Chonoles May 20 '13 at 19:26
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    $\begingroup$ I have voted your question down and to close. You are not a very new user have been told repeatedly not to ask questions in this manner, but you continue to do so. You really should stop. $\endgroup$ – tomasz May 20 '13 at 19:27
  • $\begingroup$ for part a it seems that it is trivial, so i am having a hard time trying to figure out what to show. I know that the det cannot 0. $\endgroup$ – Breezy May 20 '13 at 19:28


  1. You have $G=\text{GL}(n,\mathbb{R})$ with the operation of matrix multiplication and $H=\mathbb{R}^*$ with the operation of multiplication. You gave $\det : \text{GL}(n,\mathbb{R}) \to \mathbb{R}^*$. You need to show that composing two elements of $G$ and then moving it over to $H$ gives the same as moving two elements of $G$ over to $H$ and composing them there. Can you show that $\det(AB) = \det(A)\det(B)$?
  2. To show that $\ker \phi < G$, just need to check all of the group axioms. If $a,b \in \ker \phi$, then can you show that $ab \in \ker\phi$, i.e. if $\phi(a) = \phi(b) = 1_H$, can you show that $\phi(ab) = 1_H$? Associativity is inherited from $G$. Next you need to show that $1_G \in \ker\phi$, i.e. $\phi(1_G) = 1_H$. Finally, if $a \in \ker\phi$, can you show that $a^{-1} \in \ker\phi$?
  • $\begingroup$ 1. so $\phi(AB)=det(AB)=det(A)det(B)=\phi(A)\phi(B)$ $\endgroup$ – Breezy May 20 '13 at 20:00

For the first part, you must show that if $A\in\text{GL}_n(\Bbb R)$, then $\det(A)\in\Bbb R$ and $\det(A)\ne0$, and that for $A,B\in\text{GL}_n(\Bbb R)$, we have $\det(AB)=\det(A)\det(B)$.

You should already know of at least one way to show that a subset $K$ of a group $G$ is a subgroup of $G$. Use such a way to show that the kernel of $\phi$ is a subgroup of $G$ if $\phi:G\to H$ is a group homomorphism.


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