# Exploring Quotient Groups of D6 & D9

• zcdfhn
So, you have the 8 possible subgroups of D_6. Now, you need to find the normal subgroups among those. The normal subgroups will be: {id}, D_6, {id,r}, {id,r^2}, {id,r^3}, {id,r^4}, {id,r^5}, {id,r^6}.In summary, the possible quotient groups of D6 and D9, the dihedral group of 12 and 18 elements, are D6, D9, Z2, Z3, and Z6.

#### zcdfhn

Find, up to isomorphism, all possible quotient groups of D6 and D9, the dihedral group of 12 and 18 elements.

First of all, I don't understand the question by what they mean about "up to isomorphism." Does this mean by using the First Isomorphism Theorem? Also does this question imply that the quotient group of D6 and D9 depend on each other, or is the question asking for two different things.

"Up to isomorphism" means that if two quotient groups are different (that is, are composed of different elements) but isomorphic to each other (e.g. both are cyclic groups of order 6), then that only counts as one group.

Also, I'm pretty sure that it's two separate questions...the quotient groups of D6 and D9 are independent.

Thanks that's much clearer, so my approach to find all the quotient groups of D6 is to use the 1st isomorphism theorem, so I start with finding all the normal subgroups of D6, which are {id}, {id, R^2, R^4}, {id, R, R^2, R^3, R^4, R^5}, and D3, where R = rotation by pi/3 (I'm not sure if I'm missing anything else). Then I figure out what D3/{id}, D3/{id, R^2, R^4}, D3/{id, R, R^2, R^3, R^4, R^5}, and D3/D3, but the problem is I don't know what they are isomorphic to, especially the quotients in the middle. (I think they are isomorphic to D3, Z2, Z2, and id, respectively)

Please tell me what to fix or what I'm doing correctly, etc. Thank you.

You're making progress, but there are still flaws.

D3/{id}, D3/{id, R^2, R^4}, D3/{id, R, R^2, R^3, R^4, R^5}, and D3/D3

All those "numerators" should be D6. This will change the rest of your work.

You are missing two normal subgroups, one of order two and one of order six.

The (proper) subgroups that are quotients of D_6 can only have orders 2,3,4,6. There aren't many groups of those orders - clearly there is only one group of order 3 and one of order 2, so the order uniquely determines those. There are only groups of order 6, one abelian and one not, so those are easy to spot.