Let $H \rtimes K$ be the semidirect product of groups $H$ and $K$, not necessarily subgroups of anything.

In Theorem 10, part (5) of Dummit & Foote, Algebra, there's

Identifying $H$ and $K$ with their isomorphic copies in $H\rtimes K$, namely $\{(h,1)\}$ and $\{(1,k)\}$, we have

E) for all $h \in H$, $k \in K$, $k h k^{-1} = k \cdot h = \varphi(k)(h)$.

In other words, the hom into $Aut(H)$ associated with this semidirect product, $\varphi$, sends $k$ to conjugation by $k$ in $H \rtimes K$. Or something like that...

Their proof is:

$$ (1, k) (h, 1) (1, k)^{-1} = \\ (1 k\cdot h, k) (k^{-1} \cdot 1, k^{-1}) = \\ (k\cdot h, k)(1, k^{-1}) = \\ (k\cdot h, 1) $$

$(k\cdot h, 1) = (\varphi(k)(h), 1) = \tilde{\varphi}(k)(h)$

This is all very confusing. Is there a way to clear things up?

I'm not getting how you can mix the two isomorphisms and $\varphi$

  • 1
    $\begingroup$ Why do you change the roles of $\;H,K\;$ ? In D&F it is $\;\phi:K\to\,$ Aut$\,(H)\;$ $\endgroup$ – DonAntonio Nov 26 '13 at 20:39
  • $\begingroup$ Yes, that is correctomundo. Thank you. $\endgroup$ – Shine On You Crazy Diamond Nov 26 '13 at 21:16

By definition, we have that for $\;a,c\in H\,,\,\,b,d\in K\;$

$$(a,b)(c,d):=(a\cdot c^b,bd)$$

Note that we usually denote $\;\phi(c)(b)=:b^c\;$ and then clearly we get conjugation in the semidirect product.

Now, first

$$(h,k)\in H\rtimes K\implies (h,k)^{-1}=\left(h^{-k^{-1}}\,,\,k^{-1}\right)\;,\;\;\text{since}$$

$$(h,k)\left(h^{-k^{-1}},k^{-1}\right)=\left(h\cdot\left(h^{-k^{-1}}\right)^k\,,\,kk^{-1}\right)=(h\cdot h^{-k^{-1}k},1)=(1,1)$$

So that now

$$(1,k)(h,1)(1,k)^{-1}=(1\cdot h^k,k)(1,k^{-1})=(h^k,1)\in H\times 1$$

  • $\begingroup$ I don't understand your notation. Please use the definition $\varphi: K \to Aut(H)$, and $(a,b)(x,y) := (a \varphi(b)(x), by)$ $\endgroup$ – Shine On You Crazy Diamond Nov 26 '13 at 21:19
  • $\begingroup$ Nop, I don't use that. For me, $\;\phi(b)x:=x^b\;$ . It's the same and that's the way you can easily see that this is really conjugation in the semidirect product. Note that even D&F use the notation $\;k\cdot h\;$ which is what you call $\;\phi(k)h\;$ and I call $\;h^k\;$ . you can check the literature and you'll find that my notation is pretty widespread and used by, perhaps, msot authors. $\endgroup$ – DonAntonio Nov 26 '13 at 22:14

Let $f,g$ be the isomorphisms sending $H,K$ into $H \rtimes K$ respectively. Then, $\psi: Aut(H) \approx Aut(f(H))$ is an isomorphism given by $\phi \mapsto f \phi f^{-1}$.

Define $\tilde{\varphi} : g(K) \to Aut(f(H))$ by $\tilde{\varphi}(\tilde{k}) = f\varphi(g^{-1}(1,k))f^{-1}$.

Then $\tilde{\varphi}(k)$ on $(h,1)$ is

$$ f(\varphi(k)(h)) = (\varphi(k)(h), 1) $$

So what I think they want to say is that $\tilde{\varphi}$, the corresponding $\varphi$ for $f(H)$ and $g(K)$ is such that it's conjugation on $\tilde{H}$.


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