shakalandro
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Prove that a group of order 30 can have no more than 7 subgroups of order 5.
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- Thread starter shakalandro
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if |H|=5 is a subgroup of G then it contains 4 elements of order 5.
We can then "fit" 7 of these into 30, giving us 28 elements having order 5. Add in the identity and we have 29, leaving a single element of order 2. Its not hard to show that if any two of these intersect then one is a subgroup of the other. Being that they are of the same prime order (cyclic), that means they're the same.
That said, you can't have 7 (look at the sylow theorems), but this proves you can't have more than 7.
We can then "fit" 7 of these into 30, giving us 28 elements having order 5. Add in the identity and we have 29, leaving a single element of order 2. Its not hard to show that if any two of these intersect then one is a subgroup of the other. Being that they are of the same prime order (cyclic), that means they're the same.
That said, you can't have 7 (look at the sylow theorems), but this proves you can't have more than 7.
daon said:if |H|=5 is a subgroup of G then it contains 4 elements of order 5.
We can then "fit" 7 of these into 30, giving us 28 elements having order 5. Add in the identity and we have 29, leaving a single element of order 2.
And two elements of order 3, makes 32 elements so far.