PRACTICE PROBLEMS WITH SOLUTIONS
January 17, 2007
Here are a few practice problems on groups.
Try to work these WITHOUT looking at
the solutions!
After you write your own solution, you can compare to my solution.
Your
solution does not need to be identical—there are often many ways to solve a problem—but
it does need to be CORRECT.
1.
Suppose that
G
is a set and that
*
is an operation on
G
that satisFes the following
conditions:
a.
G
is closed under
*
.
b.
*
is associative.
c. There exists an element
e
∈
G
such that
a
*
e
=
a
for every
a
∈
G
.
d. ±or each
a
∈
G
there exists an element
b
∈
G
such that
a
*
b
=
e
.
Show that
G
is a group.
NOTE: You do NOT know that
G
is a group, that is what you are trying to prove. You
are given that SOME of the requirements of a group are satisFed, and you have to prove
that the remaining requirements are also satisFed. You do NOT know that each element has
an inverse, that is part of what you are trying to prove.
Solution
Here, and in most instances involving generic group notation, I’ll omit writing the symbol
“
*
” explicitly.
The hypotheses of this problem tell us that
most
of the properties required for
G
to be a
group are satisFed.
We just have to show that the remaining properties are also satisFed.
SpeciFcally, we have to show the following two things.
(1)
ea
=
a
for every
a
∈
G
.
(2) ±or each
a
∈
G
, the element
b
∈
G
which satisFes
ab
=
e
ALSO satisFes
ba
=
e
.
Let us work on statement (2) Frst.
Let
a
be any element of
G
.
Then we know by
assumption c that
ae
=
a
, and we know by assumption d that there is an element
b
∈
G
such
that
ab
=
e
. ±urther, by assumption d there is an element
c
such that
bc
=
e
. However, we
do NOT yet know whether
c
=
a
! We cannot assume this! Instead, using the associativity
of the operation, we compute that
ba
=
(
ba
)
e
=
b
(
ae
)
=
b
(
a
(
bc
))
=
b
((
ab
)
c
)
=
b
(
ec
)
=
(
be
)
c
=
bc
=
e.
This establishes statement (2).
Now let us prove that statement (1) is true. Let
a
∈
G
be given. We have to show that
ea
=
a
. This follows from the calculation
ea
=
(
ab
)
a
=
a
(
ba
)
=
ae
=
a.
±
1
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 Spring '10
 Staff
 Math, Algebra, Addition, ghg 1

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