zhzhzoo commented 10 years ago

1 The Definition of Group

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(1) 1 = (1 2 3), x = (2 3 1), x^2 = (2 3 1)(2 3 1) = (3 1 2), y = (2 1 3), xy = (2 3 1)(2 1 3) = (1 3 2), x^2y = (3 1 2)(2 1 3) = (3 2 1)

(2)

	1	x	x^2	y	xy	x^2y
1	1	x	x^2	y	xy	x^2y
x	x	x^2	1	xy	x^2y	y
x^2	x^2	1	x	x^2y	y	xy
y	y	x^2y	xy	1	x^2	x
xy	xy	y	x^2y	x	1	x^2
x^2y	x^2y	xy	y	x^2	x	1

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(a) Associativity: By the associativity of matrix multiplication. Identity: The identity matrix. Inverse: The inverse matrix. Every matrix in $GL_n(\mathbb{R})$ has an inverse matrix since it's determinant is not zero. Closure: The product of two invertible matrix is also invertible.

(b) Associativity: By the associativity of maps. Identity: The identity permutation. Inverse: The inverse permutation. Closure: By definition of $S_n$

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Associativity: By the the associative law of composition. Identity: The identity element. Inverse: Elements in the subset are all invertible.

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$x^{-1}w^{-1}z$

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No. A counter example is in $S_3$ : $(xy)xy=1$ but $x(xy)y=(xy)yx=x^2\not=1$ .

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((ab)c)d, (a(bc))d, a((bc)d), a(b(cd)

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(ab)c = ac = a, a(bc) = ab = a.

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$A=\left[\begin{matrix}1&0\\0&2\end{matrix}\right]$ , $B=\left[\begin{matrix}0&1\\2&0\end{matrix}\right]$

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$b=1b=a^{-1}ab=a^{-1}a=1$ , $b=1b=a^{-1}ab=a^{-1}$

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$x=1x=a^{-1}ax=a^{-1}b$

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Associativity: $(a\circ b)\circ c=c(ba)=(cb)a=a\circ(b\circ c)$ Identity: $1\circ x=x1=x,\qquad x\circ 1=1x=x$ Inverse: $x^{-1}\circ x=xx^{-1}=1,\qquad x\circ x^{-1}=x^{-1}x=1$ Closure: Inherited from G.