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I'm currently studying semi-direct product of groups (from the very famous book from Dummit & Foote). This is the context (page 176):

Let $H$ and $K$ be groups, $\varphi\in Hom(K,Aut(H))$. Let $\cdot$ denote the left action of $K$ on $H$ determined by $\varphi$. Consider $G$ the set of ordered pairs $(h,k)$ with $h\in H$ and $k\in K$ and define the following multiplication on $G$ $$(h_1,k_1)(h_2,k_2)=(h_1 (k_1\cdot h_2),k_1k_2)$$ right after that, several things are suppose to happen and the one that confuses me is this

$\textbf{(5)}$ $\forall h\in H,k\in K$, $khk^{-1}=k\cdot h=\varphi(k)(h)$

Now, I kinda understand how does the author justifies this but it uses a correspondence between the sets $H$ and $\{(h,1):h\in H\}$ and similarly for $K$. Then they say

\begin{eqnarray*} (1,k)(h,1)(1,k)^{-1}&=&(1,k)(h,1)(1,k^{-1})\\ &=&\ldots\\ &\vdots&\\ &=&(k\cdot h,1) \end{eqnarray*}

And finally says, identifying $(h,1)$ with $h$ and $(1,k)$ with $k$ we have $khk^{-1}=k\cdot h$, but I don't understand how to make the proper identification, my first shot was to use the correspondence $(h,1)\mapsto h$ but it doesn't seem to be working.

Thanks in advance for any comments

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  • $\begingroup$ I'm having difficulty understanding what you don't understand. The correspondence between $H$ and $\{(h,1):h\in H\}$ and between $K$ and $\{(1,k):k\in K\}$ is not a very deep concept. You seem to understand that $(h,1)\mapsto h$ gives you the desired bijection. So what is the problem that you are having? Do you understand the manipulations that get you from $(1,k)(h,1)(1,k)^{-1}$ to $(k\cdot h,1)$? That's the meat of this proof. $\endgroup$ – Matt Samuel Aug 29 '16 at 22:37
  • $\begingroup$ I do understand the those manipulations, what I don't understand is the jump from $(1,k)(h,1)(1,k)^{-1}=(k\cdot h,1)$ to $khk^{-1}=k\cdot h$, I'd like to think that there is some homomorphism that makes it work but I just can't see it, since the isomorphisms between $H$ and $\{(h,1):h\in H\}$ is not the same as the one between $K$ and $\{(1,k):k\in K\}$. $\endgroup$ – Cristian Baeza Aug 30 '16 at 1:29
  • $\begingroup$ Writing $khk^{-1}$ is just an abuse of notation. We're just pretending that $(h,1)$ is $h$ and $(1,k)$ is $k$. They're not. There are injective homomorphisms from $H$ and $K$ making this make sense, but they are indeed different homomorphisms from different groups. $\endgroup$ – Matt Samuel Aug 30 '16 at 2:21
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I always find it helpful to draw comparisons between a new concept and those that are more familiar to me. For example, you're probably used to writing complex numbers in the form $z = x + iy$, but this is an abuse of notation very similar to that used in your example. Consider for a moment that by definition, a complex number is a pair of real numbers $z = (x,y)$. We define addition and multiplication by the rules $(x,y) + (u,v) = (x+u,y+v)$ and $(x,y)(u,v) = (xu-yv,xv+yu)$. So how does the notation $z=x+iy$ enter the picture?

First, note that, in terms of this definition, a real number is not a complex number. However, the injective map $\mathbb{R} \longrightarrow \mathbb{C}$ given by $x \mapsto (x,0)$ allows us to embed the real numbers into the complex numbers and hence think of $\mathbb{R}$ as a subset of $\mathbb{C}$ (in fact, it's a real subalgebra!). This is precisely what Dummit and Foote do with the map $H \longrightarrow H \rtimes K$, that sends $h$ to $(h,1)$. Similarly, there's an injection $\mathbb{R} \longrightarrow \mathbb{C}$ sending $y$ to $(0,y)$ (which, btw, identifies the imaginary axis as a copy of the real line; note that $i = (0,1)$). This is analogous to the map that sends $k$ to $(1,k)$.

What about the expression $khk^{-1}$? In our example of the complex numbers, when we write $z = x + iy$, we really mean $z = (x,0) + (0,1)(y,0)$, which, if you follow the rules for complex multiplication and addition, amounts to $z = (x,y)$. Since the notation $z = x + iy$ is so terribly convenient, we allow the abuse of notation. And for similar reasons, we write $khk^{-1}$ rather than the much bulkier (though more precise) expression $(1,k)(h,1)(1,k)^{-1}$.

There's actually a second (though milder) abuse of notation used in your example. We write $k \cdot h$ instead of $\varphi(k)(h)$ because it too is shorter and easier to work with. Since $\varphi(k)(h)$ is an element of $H$, we write it as $(\varphi(k)(h), 1)$ or, combining both abuses simultaneously, simply as $k \cdot h$, rather than $(k \cdot h, 1)$.

BTW: I've never really liked the expression "abuse of notation". Software developers refer to such abuses by the much cooler name of "syntactic sugar". A little notational sugar helps the medicine go down in math too!

Anyway, cheers!

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  • $\begingroup$ $k\cdot h$ isn't an abuse of notation, that has been defined as an action of $K$ on $H$. $\endgroup$ – Matt Samuel Sep 17 '16 at 12:03

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