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Lecture

# MAT301H1 Lecture Notes - Surjective Function, Abelian Group, Cyclic Group

Department
Mathematics
Course Code
MAT301H1
Professor
all

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MAT301 ASSIGNMENT 6
DUE DATE: AUGUST 6, 2014 AT THE BEGINNING OF YOUR TUTORIAL
Question 1. (a) Show that GHis Abelian if and only if Gand Hare Abelain.
Solution. Suppose Gand Hare Abelian. Given x, y GH, then x= (g, h), y = (a, b) for some
g, a G, h, b H. Moreover,
xy = (g, h)(a, b)=(ga, hb)=(ag, bh)=(a, b)(g, h) = yx
Hence, GHis Abelian.
Conversely, suppose GHis Abelian. Let g, a G, then (g, e),(a, e)GH(where eis the identity
element in H), and
(ga, e)=(g, e)(a, e) = (a, e)(g, e)=(ag, e)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 φgInn(G), we have φg(x) = gxg1=xgg1=x=φe(x)φg=
φe(where eis the identity element in G. Hence, Inn(G) = {φe}⇒|Inn(G)|= 1.
Conversely, suppose |Inn(G)|= 1 Inn(G) = {φe}. Given g, h G, then φg, φhInn(G)φg=φe=
φhghg1=φg(h) = φe(h) = hgh =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
x
n
M
i=1
Z(Gi)x= (g1, g2, ..., gn) for some giZ(Gi)
Given y= (h1, h2, ..., hn)G,
xy = (g1, g2, ..., gn)(h1, h2, ..., hn)=(g1h1, g2h2, ..., gnhn)
= (h1g1, h2g2, ..., hngn)
= (h1, h2, ..., hn)(g1, g2, ..., gn) = yx
That is, xZ(G). Hence,
n
M
i=1
Z(Gi)!Z(G)
1