In: Advanced Math
How many subgroups of order 9 and 49 may there be in a Group of order 441
We will Sylow third theorem to find number of subgroups of order 9 and 49 in a group of order 441 which states that ,
Sylow third theorem : Supppse
be a group with
where
be a prime does not divides
If
denotes
number of subgroup of order
then ,

divides
and

Now ,
and 3 does not divides 49 .
So if
denotes number
of subgroup of order 9 then by Sylow third theorem ,
and

Now 1, 7 , 49 are all satisfies the equation
.
Hence possible number of subgroup of order
are 1 , 7 , 49
.
Also ,
and 7 does not divides 9 so by Sylow third theorem if
denotes number of subgroup of order 7 then ,
and

Now 1 satisfies
but 3 , 9 does not divides
.
Hence number of subgroup of order 49 is 1.
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If you have any doubt or need more clarification at any step please comment.