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**preview**shows page 1. to view the full**4 pages of the document.**MAT301 ASSIGNMENT 6

DUE DATE: AUGUST 6, 2014 AT THE BEGINNING OF YOUR TUTORIAL

Question 1. (a) Show that GâŠ•His Abelian if and only if Gand Hare Abelain.

Solution. Suppose Gand Hare Abelian. Given x, y âˆˆGâŠ•H, then x= (g, h), y = (a, b) for some

g, a âˆˆG, h, b âˆˆH. Moreover,

xy = (g, h)(a, b)=(ga, hb)=(ag, bh)=(a, b)(g, h) = yx

Hence, GâŠ•His Abelian.

Conversely, suppose GâŠ•His Abelian. Let g, a âˆˆG, then (g, e),(a, e)âˆˆGâŠ•H(where eis the identity

element in H), and

(ga, e)=(g, e)(a, e) = (a, e)(g, e)=(ag, e)â‡’ga =ag

Hence, Gis an Abelian group. A similar proof shows that His an Abelian group.

(b) Prove that Gis Abelain if and only if |Inn(G)|= 1.

Solution. Suppose Gis Abelian. Given Ï†gâˆˆInn(G), we have Ï†g(x) = gxgâˆ’1=xggâˆ’1=x=Ï†e(x)â‡’Ï†g=

Ï†e(where eis the identity element in G. Hence, Inn(G) = {Ï†e}â‡’|Inn(G)|= 1.

Conversely, suppose |Inn(G)|= 1 â‡’Inn(G) = {Ï†e}. Given g, h âˆˆG, then Ï†g, Ï†hâˆˆInn(G)â‡’Ï†g=Ï†e=

Ï†hâ‡’ghgâˆ’1=Ï†g(h) = Ï†e(h) = hâ‡’gh =hg. Hence, Gis Abelian.

(c) Suppose G1, G2, ..., Gnare groups. Let

G=

n

M

i=1

Gi

Prove

Z(G) =

n

M

i=1

Z(Gi)

Solution. Suppose

xâˆˆ

n

M

i=1

Z(Gi)â‡’x= (g1, g2, ..., gn) for some giâˆˆZ(Gi)

Given y= (h1, h2, ..., hn)âˆˆG,

xy = (g1, g2, ..., gn)(h1, h2, ..., hn)=(g1h1, g2h2, ..., gnhn)

= (h1g1, h2g2, ..., hngn)

= (h1, h2, ..., hn)(g1, g2, ..., gn) = yx

That is, xâˆˆZ(G). Hence,

n

M

i=1

Z(Gi)!âŠ†Z(G)

1

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