# Why is the identity component of a matrix group a subgroup?

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... – Pete L. Clark May 11 '11 at 2:17

1. Multiplying two matrices. That is, $\times : G \times G \to G$ is continuous.

2. You don't need to know that inversion is continuous (although it is by Cramer's rule). You just need to multiply the path by $A^{-1}$.

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Well, that was easy :) Thanks. Then can I ask, why is this function that takes pairs of matrices to their product continuous? – Aron Samkoff May 11 '11 at 1:35
@Aron: it suffices to show that multiplying matrices by matrices is continuous, and for that it suffices to observe that the result of multiplying two matrices is a matrix whose entries are polynomials in the entries of the matrices, and polynomials are always continuous. – Qiaochu Yuan May 11 '11 at 1:37

Hint. The path component and the connected component of this group containing the identity are one in the same.

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That isn't the issue the OP is having. – Qiaochu Yuan May 11 '11 at 1:30
A standard method ( it is a theorem, actually) for showing that s subset S of a group is a subgroup of G is showing that if a,b are in S, then the product a$b^-1$ is also in S. You can then also use the fact that continuous maps preserve connected components, i.e., if we have a continuous map f:X-->Y , and X has components $C_1$,..,$C_n$, then f(X) will have also exactly n components f($C_1$),..,f($C_n$). By continuity, then, the component of the identity will be sent to itself under the map A$B^-1$, showing that the component of the identity is a subgroup. – gary May 11 '11 at 1:56