Assume that $B$ is generated by $C=\{a,b\}$ where $ord(a)=5$, $ord(b)=4$ and $ab=ba^2$.
a) Prove that for any $c \in B$, $c \neq b$: $<c> \cap <b>=\{e\}$. ($<c>$ is cyclic group generated by c)
b) Any element of B can be written as $a^ib^j$ for some i,j and such representation is unique.
2nd task is intuitively obvious for me, because we can translate any element $b^ja^i$ via assumed equality to some $a^lb^k$ and such representation will be unique (but I am not quite sure how to prove that).
But I am stuck with the 1st one. I tried to do it indirectly, and we have $\{e,b^2\}$ or whole $<b>$ in such intersection, but I don't know what to do next.
I would appriciate any help or advice. Thanks in advance!
The first statement is far from correct. Just take $c = b^{2}$ or $c = b^{3}$ and see what happens.
As to the second one (which is also incorrectly stated, see below), since $B$ is generated by $a, b$, it will contain all expressions of the form $a^{i} b^{j}$. If you multiply two of these, you get $$ a^{i} b^{j} a^{s} b^{t} = a^{i} b^{j} a^{s} b^{-j} b^{j + t} = a^{i + 3 s} b^{j + t}, $$ where $$ b^{j} a^{s} b^{-j} = a^{3 s} $$ derives from the relation $b^{-1} a b = a^{2}$, in the equivalent form $b a b^{-1} = a^{3}$.
So $B$ consists indeed of the elements $a^{i} b^{j}$, with $0 \le i < 5$ and $0 \le j < 4$.
To see they are distinct, you may note that these relations are satisfied in the group of order $20$ generated by the permutations $a = (12345)$ and $b = (2354)$.