Cfrgtkky

I want to solve the following problem from Dummit & Foote's Abstract Algebra text (p. 184 Exercise 6):

Assume that $K$ is a cyclic group, $H$ is an arbitrary group and $varphi_1$ and $varphi_2$ are homomorphisms from $K$ into $text{Aut}(H)$ such that $varphi_1(K)$ and $varphi_2(K)$ are subgroups of $text{Aut}(H)$. If $K$ is infinite assume $varphi_1$ and $varphi_2$ are injective. Prove by constructing an explicit isomorphism that $H rtimes_{varphi_1} K cong H rtimes_{varphi_2} K$ (in particular, if the subgroups $varphi_1(K)$ and $varphi_2(K)$ are equal in $text{Aut}(H)$, then the resulting semidirect products are isomorphic). [Suppose $sigma varphi_1(K) sigma^{-1}=varphi_2(K)$ so that for some $a in mathbb{Z}$ we have $sigma varphi_1(k) sigma^{-1}=varphi_2(k)^a$ for all $k in K$. Show that the map $psi:H rtimes_{varphi_1} K to H rtimes_{varphi_2} K$ defined by $psi((h,k))=(sigma(h),k^a)$ is a homomorphism. Show $psi$ is bijective by constructing a 2-sided inverse.]

My attempt:
Let $K=langle x rangle$, and suppose $sigma in text{Aut}(H)$ conjugates $varphi_1(K)$ to $varphi_2(K)$, that is

$$sigma varphi_1(K) sigma^{-1}=varphi_2(K) tag{1}.$$

From this we find
$$sigma varphi_1(x) sigma^{-1}=varphi_2(x^a) tag{2} $$
for some $a in mathbb{Z}$. Since all elements of $K$ are powers $x^n$ of $x$, raising this equality to the $n$-th power gives
$$forall k in K:sigma varphi_1(k) sigma^{-1}=varphi_2(k^a)=varphi_2(k)^a tag{3}.$$

We now prove that the suggested $psi$ is a group homomorphism:

begin{equation}
begin{split}
psi((h_1,k_1)(h_2,k_2))&=psi(h_1 varphi_1(k_1)(h_2),k_1k_2)=(sigma(h_1 varphi_1(k_1)(h_2)),(k_1k_2)^a)\
&=(sigma(h_1)(sigma varphi_1(k_1))(h_2),k_1^a k_2^a)=(sigma(h_1)(varphi_2(k_1^a) sigma)(h_2),k_1^ak_2^a)\
&=(sigma(h_1),k_1^a)(sigma(h_2),k_2^a)=psi((h_1,k_1))psi((h_2,k_2))
end{split}
end{equation}

Where we have used the fact that $K$ is abelian, alongside with the homomorphism law for $sigma$ and equation $(3)$. We're left with showing that $psi$ has a 2-sided inverse, which will be done in two cases:

• Assume $K=langle x rangle$ is infinite cyclic. Property $(3)$ gives
  $$varphi_2(K)=varphi_2(K^a) tag{4}$$
  where $K^a$ is the image of $K$ under the $a$-th power homomorphism $K to K:k mapsto k^a$. Since $varphi_2$ is injective, this is only possible if the $a$-th power homomorphism is surjective, which happens iff $a=pm 1$. We can thus see that
  $$chi((h,k))=(sigma^{-1}(h),k^a)$$
  is a 2-sided inverse of $psi$.



• 
  

  Assume $K=langle x rangle cong Z_n$ is finite cyclic of order $n$, and denote the orders of the cyclic groups $varphi_1(K),varphi_2(K)$ by $m$. I believe that $(a,n) = 1$ so that there is some integer $b$ such that $ab equiv 1 pmod{n}$. If we have this, we can see that
  $$chi((h,k))=(sigma^{-1}(h),k^b) $$
  is a 2-sided inverse of $psi$. However, all I could show is the following.

  
  
  
  
  
  1. Obviously, $m|n$.
  
  
  2. Since $varphi_1(K)=langle varphi_1(x) rangle$ and $varphi_2(K)=langle varphi_2(K) rangle$ we have $|varphi_1(x)|=|varphi_2(x)|$. According to equation (3) $varphi_2(K)$ is also generated by $varphi_2(x^a)$, so that $|varphi_2(x^a)|=|varphi_2(x)|$ which gives $(a,m)=1$.
  
  

  3. Raising equation $(2)$ to the power of $|x^a|=frac{n}{(a,n)}$ gives $1=varphi_2(1)=varphi_2((x^{a})^{frac{n}{(a,n)}})=varphi_2(x^a)^{frac{n}{(a,n)}}$, so that $m| frac{n}{(a,n)}$.

  
  

  4. From the first occurrence of "$a$" in equation $(2)$, and the fact that $varphi_2(x^a)=varphi_2(x)^a$ we can see that $a$ may shifted by any multiple of $m$.

  
  
  
  

My questions:

1. Are there any flaws with my proof?




2. It seems that in the infinite case, it is only necessary to assume one of the $varphi_i$'s to be injective. Is this true?




3. Is it possible to prove that "$a$" can be chosen coprime with $n$?

Thank you!

If $m$ is a divisor of $n$ then the natural map $mathbb{Z}_nrightarrowmathbb{Z}_m$ induces a surjective homomorphism $U(n)rightarrow U(m)$, see here for a proof. You have already noticed that $(a,m)=1$. By the aforementioned fact, there exists $a'$ such that $a'equiv a;(mathrm{mod};m)$ and $(a',n)=1$. Notice $varphi_2(k)^a=varphi_2(k)^{a'+lm}=varphi_2(k)^{a'}$. Therefore by replacing $a$ with $a'$ you can assume that $(a,n)=1$.