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\begin{center}
{{\Large\bf MODERN ALGEBRA II}\\
\medskip

{\Large\bf HOMEWORK ASSIGNMENT VI}\\
\medskip
{\bf (Turn in underlined problems.)}}
\end{center}

\vspace{.5in}

\noindent
\underline{\large\bf ADDITIONAL PROBLEMS}
\vspace{.25in}

{\bf 

The purpose of these exercises is to determine 
the structure of a group $G$ that has 
complementary subgroups
$H, N < G$ with $N\lhd G$
as in the following picture:

\bigskip

\bigskip

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\bigskip

\bigskip

(Example: $G = S_n, N = A_n, H = \{1, (1\;2)\}$.)

\bigskip

1. 
Suppose we have groups $H$ and $N$ and a homomorphism
$\varphi:H\rightarrow {\rm Aut}(N)^{op}$. For $n\in N$
and $h\in H$ write $n^h$ to denote $(\varphi(h))(n)$.
Define a multiplication on the set $H\times N$ as follows:
\[(h_1,n_1)*(h_2,n_2) = (h_1\cdot h_2,n_1^{h_2}\cdot n_2).\]

\noindent
\underline{(a)} 
Show that this is an associative multiplication.
(You will need to explain why $(n^h)^k = n^{(hk)}$.
Why is important to use ${\rm Aut}(N)^{op}$ 
instead of ${\rm Aut}(N)$?)

\noindent
\underline{(b)} 
Show that there is an identity element for this
multiplication, and that each element has
a multiplicative inverse.

\smallskip

The set $H\times N$ under these group operations 
is called the {\em semidirect product} of $H$ and $N$,
and is denoted $H\times_{\varphi} N$.

\medskip

\underline{2.} Show that the subset of 
$G = H\times_{\varphi} N$
of elements of the form $(h,1)$ comprise a subgroup
$H' < G$ isomorphic to $H$.
Show that the subset of $G$
of elements of the form $(1,n)$ comprise a {\it normal}
subgroup $N'\lhd G$ isomorphic to $N$. 
Show that $H'$ and $N'$ are complements in 
the subgroup lattice of $G$.

\medskip

The previous two exercises show that the semidirect product 
$G = H\times_{\varphi} N$ has a subgroup lattice 
like the one
pictured. Now we prove that the converse is true:
a group with this type of subgroup lattice is 
isomorphic to a semidirect
product.

\underline{3.} 
Assume that $H, N < G$ and that
$N\lhd G$. For $h\in H$ define a function 
${}^h:N\rightarrow N: n\mapsto h^{-1}nh$.
That is, $n^h = h^{-1}nh$. Show that 
this function is an automorphism of $N$.

\medskip

\underline{4.} Continuing the previous problem, 
show that the function 
$\varphi : H\rightarrow {\rm Aut}(N)^{op}: h\mapsto {}^h$
is a homomorphism.

\medskip

5. Suppose that $G$ has complementary subgroups 
$H$ and $N$ with $N\lhd G$. Using $\varphi$
as defined in the previous problem, show that
$G\cong H\times_{\varphi} N$.
(Hint: Show that the function 
$f((h,n)) = hn$
is an isomorphism 
from $H\times_{\varphi}N$ to $G$.)

\medskip

6. Show that $H\times_{\varphi} N = H\times N$
when $\varphi:H\rightarrow {\rm Aut}(N)^{op}$ is the
constant function $\varphi(x) = 1$.
(A direct product is a special semidirect
product.)

\medskip

For Graduate Credit:

F. Show that $A_4\cong {\BZ}_3\times_{\varphi} 
({\BZ}_2\times {\BZ}_2)$ for some $\varphi$.

\bigskip

\noindent
{\bf Question:} Can you find any group that 
cannot be constructed from cyclic groups
using the semidirect product construction
repeatedly?


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