Help with group theory problem please I have a problem and my solution to parts $(a)$, $(b)$, and $(c)$ of $(a), (b), (c), (d)$. I want to know if what I have done so far is correct and how I can proceed further.
Definition: A group $G$ is called a "hanuman" if every element of $G$ has order $2$ or $1$. 
Lemma: Every hanuman is abelian. Proof: Let G be a hanuman. Then $gg=1$ for all $g$ in $G$, which means every element is its own inverse. Thus, for all $a$ and $b$ in $G$, we have $ab=(ab)^{-1}=b^{-1}a^{-1}=ba$.
Part (a): Show that every hanuman with more than one element has a normal subgroup $H$ such that $H = 2$.
My solution: Case 1: $o(h)=1$. Then $\langle h \rangle$= ${h}$. Thus, $o(h)=1$ is not what we want. Case 2: $o(h)=2$. Then $\langle h \rangle$ = $[h,h^2]=[h,1]$. Thus $|\langle h \rangle| = 2$.
Part (b): Show that $G/N$, where $N$ is any normal subgroup of $G$, is also a hanuman.
My solution: Let $a$ be in $G/H$. Then $a=gH$ for some $g$ in $G$. Thus $a^{-1}=(gH)^{-1}=g^{-1}H=gH$. Hence $a$ is its own inverse. So, $G/H$ is also a hanuman.
Part (c): Deduce from parts (a) and (b) that: Given any finite hanuman $G$ with $|G|$ at least $2$, there exists another hanuman of cardinality $|G|/2$.
My solution: We have: $|G/H| = [G:H] \implies |G|=|H|[G:H]=2|G/H|$. Hence, $|G|/2=|G/H|$. By part (b), we have $G/H$ is a hanuman. Therefore $G/H$ is a hanuman with $|G/H| = |G|/2.$
Part (d): Show that if $G$ is a finite hanuman, then $|G|=2^k$ for some nonnegative integer $k$.
My solution: Suppose $G$ is a hanuman with $|G|=2^km$, where $m$ is odd and $m>1$. I need to proceed to find a contradiction, but I don't know how.
Appreciate the help!

Nice creativity with the terminology here!
Seriously, your work is largely excellent, up to part $(d)$. Suggestion for proceeding with $(d)$: make use of the result of your proof for $(c)$, applying it $k$ times.
Feel free to post a follow-up with your progress on $(d)$.
