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Math 3124
Tuesday, October 11
October 11, Ungraded Homework
The starting position is
a
b
c
d
e
a
b
e
c
d
e
d
We are given
bd
=
e
and
ec
=
d
. Now which element is the
identity? The equality
bd
=
e
rules out
b
and
d
, and the equality
ec
=
d
rules out
e
and
c
.
We are only left with
a
, so
a
is the identity. Therefore we can ﬁll in the
a
column and
a
row
to obtain
a
b
c
d
e
a
a
b
c
d
e
b
b
e
c
c
d
d
e
e
d
Now consider
bc
. By looking at the
b
row it cannot be
b
or
e
,
and by looking at the
c
column it can’t be
c
or
d
. Therefore it must be
a
. Since
bc
=
a
and
a
is the identity, we conclude that
cb
=
a
. We now consider
cd
. By looking at the
d
column it
can’t be
d
or
e
, and by looking at the
c
row it can’t be
c
or
a
. Therefore it must be
b
, and we
obtain the following table.
a
b
c
d
e
a
a
b
c
d
e
b
b
a
e
c
c
a
b
d
d
e
e
d
We must now have
c
2
=
e
, and the rest is easy; the ﬁnal table is
a
b
c
d
e
a
a
b
c
d
e
b
b
d
a
e
c
c
c
a
e
b
d
d
d
e
b
c
a
e
e
c
d
a
b
Problem 17.15 on page 92. Let
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This note was uploaded on 01/02/2012 for the course MATH 3124 taught by Professor Parry during the Fall '08 term at Virginia Tech.
 Fall '08
 PARRY
 Math, Algebra

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