Correction for some group problems So I just had my final exam, and I got marked down for this problem which goes:
Let $R$ be a ring with unity $1$,  $a\in R$ satisfies $a^7=0$. 
i) Prove that $H = \{1+a^2xa^5| x \in R\}$ is a group.
ii) Assume $R$ is finite, prove that $|H|$ is a divisor of $|R|$.
i) I got marked down because I said the inverse is $(1+(a^2(-x)a^5+.....+(a^2(-x)a^5)^6)$, but why is this wrong because, $(1-a^2(-x)a^5)(1+(a^2(-x)a^5)+.....+(a^2(-x)a^5)^6)=1-(a^2 x a^5)^7=1-a^2xa^5*a^2xa^5*.... = 1-0=1$
Since $a^7$ is $0$. Can we not do this?
ii) I got marked down for saying there is a well defined bijection $f(x) =1+a^2xa^5$ from $N_a$ to $H$ where $N_a = \{a^2xa^5| x\in R\}$, because he said I should have written $f(a^2xa^5)=1+a^2xa^5$ instead which I do understand but just need alittle further clarification to why this is so.
I accept my mistake partly in ii), but I fail to see the fault in the inverse I gave for part i). Could anyone please clarify for me the mistakes I have made? 
 A: Let $f(x) = 1- a^2 x a^5$. We want to show that $H=\{f(x) : x \in R\}$ constitutes a group. One question I had is whether this is under addition or multiplication, though presumably it is under multiplication inherited from $R$.


*

*$H$ has multiplicative identity, since $f(0) = 1 - a^20a^5 = 1-0 = 1$.

*Each element may have a multiplicative inverse. Searching for intuition, if we take $f(x)f(y)$, we get $(1-a^2xa^5)(1-a^2ya^5) = 1 -a^2ya^5 -a^2xa^5 + a^2xa^5a^2ya^5 = 1 - a^2ya^5 - a^2xa^5 + 0 =  1 -a^2(x+y)a^5$ (by distributivity in $R$).

*Incidentally, the fact that $f(x)f(y) = f(x+y)$ shows that $f(R)$ is closed under multiplication.

*It also shows that $f(x)f(-x) = 1$. Hence if $f(x)$ is an element of $H$, then $f(-x) = 1-a^2 (-x)a^5$ is its multiplicative inverse in $H$.  



As for part (ii), you've got the right strategy showing that the set $\{f(x) : x \in R\}$ is bijective with a normal subgroup $N_a$. I think your instructor is saying that in order to show that these two sets are bijective, you must make a function $g$ that goes betwen them. That would be a function sending 
$f(R)$ to $N_a$. As defined, however, your function seems to be more approriate for sending $R$ itself to $N_a$, and is not bijective because if $g : x \mapsto a^2 x a^5$, then $g(0)=1 = g(1)$.
You'd actually want a function such as 
$$h: N_a \rightarrow H, \qquad y \mapsto y+1 $$
