Determine all elements of the coset 3+⟨8⟩ in the group ℤ10 The question is:               
coset 3+⟨8⟩
Determine all elements of the coset 3+⟨8⟩ in the group ℤ10. 
Enter your answer as a comma separated list; make sure that each element you enter is ≥0 and <10.
I firstly solved <8>=<8,64,512,4096...> mod 10=<8,4,2,6>
Does the coset 3+<8> simply mean <3+8,3+4,3+2,3+6>=<11,7,5,9>?
Do I take mod 10 next? Is it <1,7,5,9>? 
And my one more question is:
Determine all elements of the coset 1⟨51⟩ in the group U(56).
Is the first step finding <51> and then multiply the set <51> by 1 like the above question?
Thank you!
 A: I think there is a little confusion in what you have written in your question
$10\mathbb{Z} = \{\ldots , -20, -10, 0 , 10 , 20, \ldots \}$. Now $\mathbb{Z}_{10}$ is the set of (left) cosets of $10\mathbb{Z}$ in $\mathbb{Z}$. So for example $8+10\mathbb{Z}=\{\ldots , 8-20, 8-10, 8+0 , 8+10 , 8+20, \ldots \} =\{\ldots , -12, -2, 8 , 18 , 28, \ldots \} $. Now in the group $\mathbb{Z}_{10}$ elements are (left) cosets (which are sets of integers). So it is NOT defined how to add a single integer to a coset in $\mathbb{Z}_{10}$ since individual integers  are not elements of $\mathbb{Z}_{10}$. However you can add two cosets together in $\mathbb{Z}_{10}$ to get another coset in $\mathbb{Z}_{10}$. For example $(8+10\mathbb{Z})+(3+10\mathbb{Z})$ which is defined to be the coset $(8+3)+10\mathbb{Z}$ which is $11+ 10\mathbb{Z}$. And if you work that set out you will find you get $\{\ldots, -9,1,11, \ldots \}$. (This is the same coset as $1+10\mathbb{Z}$, and in general $(a+10\mathbb{Z})+(b+10\mathbb{Z})=  c+10\mathbb{Z}$ where $c$ is the remainder of $a+b$ modulo 10).
I hope this helps  
A: You have some of the right idea (aside from some small mistakes); you are doing all computations modulo 10. We have that $\langle 8 \rangle$ denotes the (additive) subgroup generated by $8$, as a subset 
$$\langle 8 \rangle = \{8,\ 6,\ 4,\ 2,\ 0 \}$$ 
(in this context I would think of this as $\{8, 8+8=6,\ 8+8+8=4,\ \ldots\}$).  Now you are looking at the coset $3 + \langle 8 \rangle$, which consists of all elements of the form $3+a$, where $a$ is an element of $\langle 8 \rangle$.  You also want to reduce these all modulo 10, so you get 
$$3 + \langle 8 \rangle = \{ 1, \ 9,\ 7, \ 5, \ 3\}.$$
Note that you don't want your answer in angle brackets; these suggest that you are forming the subgroup generated by the elements listed inside (in this case your final result is a coset and not a group).
