these are subgroups generated by the said element.

so <a> = {1 = a^0, a^1, a^2 , a^3, ... } = {1, a , a*a, a*a*a, ...}

do that until you exhaust the table (of course you have to fill it out first).

for things like <b,e>, this is the subgroup generated by all products of b and e.

3. Write the elements of G as products of two fixed generating elements (you may use

1 to denote the identity). i think they want you to write G = <x,y> for some x,y in G. And do this twice. Perhaps this will become clearer after doing part 2.

G,*) is not isomorphic to D4 [Hint: question 1, order of elements] or

to Z/4Z × Z/2Z. tell me what D_4 is and i might be able to help.

b>is normal, and write out the left cosets of this subgroup. to show <b> is normal, we must show that $\displaystyle gbg^{-1} \in <b>$ for all $\displaystyle g \in G$. just go through the table and verify that this holds.

the left cosets will be:

<b> = {1, b, b^2, b^3, ... }

a*<b> = {a,ab,ab^2,ab^3, ...}

c*<b> = { ...}

and so on. do this until you exhaust all the elements of G

Then give

the group table for

to do this, you must pick one element out of each of the sets above and make a table with them similar to how they did for G. note that when you take the product of your representatives you may get something that is not a representative. In that case, choose the representative that is in the coset of the element you get. use [ ] brackets to denote your representative. like, use [1] to be the representative for the first coset i have above, this represents the set {1,b,b^2,b^3,...}

6. Give an isomorphism from

G/<b>to either Z/4Z or Z/2Z × Z/2Z. In the othercase, explain why no isomorphism exists. i will have to think about this more. I would probably need the answers to the previous questions to answer this, and I am too lazy to do them myself now

Sorry I could not be of much help. I just started (a basic course in) abstract algebra myself