Math 4107 Midterm 1 Solutions, Fall 2009
October 18, 2009
1.
You can look up these definitions yourself.
2.
There are many ways to solve this problem. We present two here:
Way 1.
Let
H
be the subgroup of order 35.
By Lagrange, we know that
every element of
H
has order either 1
,
5
,
7 or 35. If some element has order
5 or 35, we are done (if
g
has order 35, then
g
7
has order 5).
So, let us assume every element has order 1 or 7. So, since the identity
is the only element of order 1, every nonidentity element has order 7. If
g
1
and
g
2
are two elements of order 7, then either the cyclic group (
g
1
) = (
g
2
)
or (
g
1
)
∩
(
g
2
) =
{
e
}
.
What this means is that the subgroups of order 7
are essentially disjoint, apart from intersecting in the identity; so, letting
A
1
, ..., A
k
be all these different subgroups of order 7, together they contribute
6
k
elements of order 7, and 1 element of order 1. But this implies
35 =

H

= 1 + 6
k,
which cannot hold because 35

1 = 34 is not divisible by 3.
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 Fall '08
 Staff
 Math, Algebra, Subgroup, Cyclic group

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