About Properties of Semi-direct Product 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   
 A: 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!
