Order of elements of a quotient group

Question

Let $A$ be the (normal) subgroup of $B$. $A$ is generated by all elements $x^7$ with $x\in B$. Show that every element in $B/A$ has order $1$ or $7$. I do not really have a clue where to start or if this is even true. Could someone help?

Answer 1

Any element in the quotient group $B/A$ is of the form $xA$ such that $x\in B-A$. Now consider $(xA)^7=x^7A$, by the definition of $A$, $x^7 \in A$, therefore $(xA)^7=A$. Note that $A$ is the identity of the quotient group $B/A$. Thus order of $xA$ is a divisor of $7$, so it is either $1$ or $7$.

Answer 2

Hint: Consider what are the elements of $B/A$.

Next consider what happens to those elements if you take their $7$th power.

Last consider that $7$ is a prime number.