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I'm working through Stillwell's "Naive Lie Theory". I'm supposed to show that the identity component of a matrix group is a subgroup in two steps. I'm allowed to assume that "matrix multiplication is a continuous operation". First question- what does this mean? Does this mean multiplying matrices by a fixed matrix is continuous, or multiplying two matrices which vary?

In the first step, I'm supposed to prove that if there are continuous paths in the group $G$ from 1 to $A \in G$ and to $B \in G$ then there is a path in G from $A$ to $AB$.

I did this by assuming that matrix multiplication by a fixed matrix was continuous. I presume that this will get us closure under group operation by concatenating the path from 1 to $A$ with the path from $A$ to $AB$.

Second, and where I am stuck, is in proving that if there is a continuous path in $G$ from 1 to $A$ there is also a continuous path from $A^{-1}$ to 1. If I knew that the map that sends $A$ to $A^{-1}$ was continuous, I think I would be done, but I don't know how to get this easily.

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    Just a comment: it is not hard to show that the connected component of the identity of any topological group is a normal subgroup. It sounds like Stillwell is taking a more "naive" perspective than this, but this much naivete may not be to everyone's taste...2011-05-11

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