Insights Blog
-- Browse All Articles --
Physics Articles
Physics Tutorials
Physics Guides
Physics FAQ
Math Articles
Math Tutorials
Math Guides
Math FAQ
Education Articles
Education Guides
Bio/Chem Articles
Technology Guides
Computer Science Tutorials
Forums
General Math
Calculus
Differential Equations
Topology and Analysis
Linear and Abstract Algebra
Differential Geometry
Set Theory, Logic, Probability, Statistics
MATLAB, Maple, Mathematica, LaTeX
Trending
Featured Threads
Log in
Register
What's new
Search
Search
Search titles only
By:
General Math
Calculus
Differential Equations
Topology and Analysis
Linear and Abstract Algebra
Differential Geometry
Set Theory, Logic, Probability, Statistics
MATLAB, Maple, Mathematica, LaTeX
Menu
Log in
Register
Navigation
More options
Contact us
Close Menu
JavaScript is disabled. For a better experience, please enable JavaScript in your browser before proceeding.
You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an
alternative browser
.
Forums
Mathematics
Linear and Abstract Algebra
What is the group action of G on itself by left conjugation?
Reply to thread
Message
[QUOTE="Mr Davis 97, post: 5729829, member: 515461"] My textbook says the following: "Let ##G## be a group and ##G## act on itself by left conjugation, so each ##g \in G## maps ##G## to ##G## by ##x \mapsto gxg^{-1}##". I am confused by the wording of this. ##g## itself is not a function, so how does it map anything at all? I am assuming this is supposed to mean that ##g \cdot x = gxg^{-1}## is the definition of the action, but why does it use the map notation as if ##g## is a function, when ##g## is really an element? [/QUOTE]
Insert quotes…
Post reply
Forums
Mathematics
Linear and Abstract Algebra
What is the group action of G on itself by left conjugation?
Back
Top