Order of a Group

Question

Find the order of a group H, given the following:

(1) H is a subgroup of some group G with order 168

(2) H is a subgroup of another group K with order 112

(3) H is not cyclic and dihedral

(4) H contains an element of order 7

(5) H has more than 2 left cosets in K

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My work,

According to Lagranage theorem, we can conclude the following from (1) and (2)

From (1), since order of H divides the order of G(168), order of H can be one out of 1,2,3,4,7,8,12,14,21,24,32,56,64,168

From (2), since order of H divides the order of K(112), order of H can be one out of 1,2,4,7,8,16,28,56,112

From (1) & (2), order of H should be one out of 1,2,4,7,8,56

From (3): since H is not cyclic, it's order cannot be a prime number. Hence order of H cannot be 2 or 7 and now we're left with one out of 1,4,8,56 for order of H

How do I proceed further from here?

Answer 1

Dear Ashley,

I hope that you are doing well! I understand it is a little past time to answer this question but I will go ahead and follow up on your work. You've done a great job noticing that by Lagrange's Theorem we need some integer that divides both 168 and 112, as well as noticing that the becauase the group is not cyclic it cannot be a prime integer. We can then use the fact that each element of the group H must divide the order of the group (by another corollary of the Lagrange Theorem). Thus our only answer can be that H has an order of 56, since the group has an element of order 7. Now I shall leave it to you to make sure that it all lines up, if you need more help or if you notice an error, feel free to respond!

Sincerely,

Landon Holley

P.S. I apologize in advance if the group of H having an order of 56 breaks the dihedral statement or any of the prior statements or if it breaks the last statement, I have not worked too much with topics in group theory.