Let C8 be the cyclic group of order eight. Let G=(C8)^10 and H=(C8)^3 and H is a subgroup of G.

dead slug

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How many exist subgroups F in G such that F=(C8)^5 and H∩F is C8?
I yet understood that there are 2⋅8^(i−1)−4^(i−1) ways to choose C8 from (C8)^i. But I don't know what I can do next.
 

Jomo

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How many exist subgroups F in G such that F=(C8)^5 and H∩F is C8?
I yet understood that there are 2⋅8^(i−1)−4^(i−1) ways to choose C8 from (C8)^i. But I don't know what I can do next.
So C8 is cyclic group of order 8. What does that mean to you?
F=(C8)^10 and H=(C8)^3 . What does that mean to you?
H is a subgroup of G.
What does that mean to you?
H∩F is C8 ...
LaGrange's theorem states that ......?
 

dead slug

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Here I used simbol = as isomorphism. And I used (C8)^3 as C8(+)C8(+)C8. And also you write "F=(C8)^10" but it is not true. F=(C8)^5 and G=(C8)^10.
 

dead slug

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So C8 is cyclic group of order 8. What does that mean to you?
F=(C8)^10 and H=(C8)^3 . What does that mean to you?
H is a subgroup of G.
What does that mean to you?
H∩F is C8 ...
LaGrange's theorem states that ......?
I know that here firstly need to choose H<intersect>F in H. I can do this by 2*8^(3-1)-4^(3-1)=112 ways. So, answer is diveding by 112.
 
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