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!

  • $\begingroup$ .. I think in (4), "(3) give $\varphi_2(K) = \varphi_2(K^a)$" need some more arguments.. But, $ \varphi_2(K^a) \subset \varphi_2(K) $ is possible. $\endgroup$ – hew Oct 29 '19 at 7:55
  • $\begingroup$ It may be like using a sledgehammer to crack a peanut, but 3 can be answered by Dirichlet's theorem. $\endgroup$ – ashpool Jul 7 '20 at 22:12

If $m$ is a divisor of $n$ then the natural map $\mathbb{Z}_n\rightarrow\mathbb{Z}_m$ 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 $(a,n)=1$.

  • $\begingroup$ I can't understand why there exists $a'$ such that $(a',n)=1$. Could you explain me more precisely? $\endgroup$ – Pearl Dec 2 '18 at 22:56
  • $\begingroup$ @Pearl: $(a,m)=1$ therefore $a\in U(m)$. Since the map $U(n)\rightarrow U(m)$ is surjective we can take a preimage of $a$, call it $a'$. Since $a'\in U(n)$ then $(a',n)=1$. By the way the map $U(n)\rightarrow U(m)$ is defined we also have $a\equiv a'\;(m)$. $\endgroup$ – Mec Dec 5 '18 at 23:17

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