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The group of all non-singular n×n matrices is known as the general linear group GL(n) and that with only real elements as GL(n,R). If {R}^{∗} denotes the multiplicative group of non-zero real numbers, prove that the mapping Φ : GL(n,R) →{R}^{∗}, defined by Φ(M) = det M, is a homomorphism.
Show that the kernel K of Φ is a subgroup of GL(n,R). Determine its cosets and show that they themselves form a group.

Step-by-step

If P and Q are two matrices belonging to GL(n,R) then, under Φ,
(PQ)^{\prime}= |PQ| = |P| |Q| ={P}^{\prime}{Q}^{\prime}.
Thus Φ is a homomorphism.
The kernel K of the mapping consists of all matrices in GL(n,R) that map onto
the identity in {R}^{∗}, i.e all matrices whose determinant is 1.
To determine whether K is a subgroup of the general linear group, let X and Y
belong to K. Then, testing K for the four group-dedefining properties, we have
(i) (XY )^{\prime} = {X}^{\prime}{Y}^{\prime} = 1×1 = 1, i.e. XY also belongs to K, showing the closure of the kernel.
(ii) The associative law holds for the elements of K since it does so for all elements of GL(n,R).
(iii) | I | = 1 and so I belongs to K.
(iv) Since {X}^{−1}X = I, it follows that |{X}^{−1}| |X| = | I | and |{X}^{−1}| × 1 = 1. Hence |{X}^{−1}| = 1 and so {X}^{−1} also belongs to K.
This completes the proof that K is a subgroup of GL(n,R).
Two matrices P and Q in GL(n,R) belong to the same coset of K if
Q = PX, where X is some element in K.
It then follows that
|Q| = |P| |X| = |P| × 1.
Thus the requirement for two matrices to be in the same coset is that they have equal determinants.
Let us denote by {C}_{i} the (infinite) set of all matrices whose determinant has the value i; the label i will itself take on an infinite continuum of values, excluding 0. Then,
(i) For any {M}_{i} ∈ {C}_{i} and any {M}_{j} ∈ {C}_{j} we have
|{M}_{i}{M}_{j} | = |{M}_{i}| |{M}_{j} | = i × j,
implying that we always have {M}_{i}{M}_{j} ∈ {C}_{(i×j)} . Thus the set of cosets is closed, with {C}_{i} × {C}_{j} = {C}_{(i×j)} .
(ii) The associative law holds, since it does so for matrix multiplication in general, and the product of three matrices, and hence its determinant, is independent of the order in which the individual multiplications are carried out.
(iii) Since
|{M}_{i}{M}_{1}| = |{M}_{i}| |{M}_{1}| = i,
{C}_{i} × {C}_{1} = {C}_{i} , showing that {C}_{1} is an identity element in the set.
(iv) Since
|{M}_{i}{M}_{1/i}| = |{M}_{i}| |{M}_{1/i}| = i × (1/i) = 1,
{C}_{i} × {C}_{1/i} = {C}_{1}, showing that the set of cosets contains an inverse for each coset.
This completes the proof that the cosets themselves form a group under coset multiplication (and also that K is a normal subgroup).

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