# General Linear Groups with Homomorphisms [closed]

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, AmzotiMay 20 '13 at 20:20

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• What have you tried? (The first should be a well-known thing from linear algebra and the second is just applying the definition). – Tobias Kildetoft May 20 '13 at 19:25
• 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. – tomasz May 20 '13 at 19:27
• 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. – 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 $a^{-1} \in \ker\phi$?
• 1. so $\phi(AB)=det(AB)=det(A)det(B)=\phi(A)\phi(B)$ – 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.