Example of not so simple group ??

Help me with an example of a group having subgroups but it doesn't admit a normal subgroup.. ??

The alternate definition of simple groups using non trivial homomorphic image. Searching for a map that doesn't have a non trivial kernal ??

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Every group admits two normal subgroups, $1$ and itself. Now, as nontrivial normal subgroups goes, that's another story. –  Pedro Tamaroff Apr 28 '14 at 3:11
Better to not try to be clever in the title - I expected that you wanted something other than a simple group. BE direct, particular in forums where there are lots of foreign readers. –  Thomas Andrews Apr 28 '14 at 3:31

One classic example is $A_5$, which has plenty of non-trivial subgroups, none of which are normal. In fact, this is the smallest suitable example.

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All the other examples are of finite simple groups. However, there do exist infinite simple groups. One of the best known is Richard Thompson's group $T$. It contains a very interesting group, called $F$, as a subgroup. I'll just define these groups here. If you want a more complete introduction then look at this ams "What is..." article by J. W. Cannon and W. J. Floyd. If you want a proof that this group is simple then you should look up the article "Introductory notes on Richard Thompson's groups" by Cannon, Floyd and Parry. These notes of Cannon, Floyd and Parry are the standard reference for Richard Thompson's groups $F$, $T$ and $V$.

There are three groups introduced by Thompson in 1965, $F<T<V$, and I believe he did not publish his constructions although he did prove that $T$ and $V$ are both simple. I'll define $F$, and use this to define $T$. The definition of $V$ is more complicated so I'll leave you to look it up in one of the above references.

The group $F$: The group $F$ is the set consisting of piecewise linear homeomorphisms from the closed unit interval to itself which are differentiable at all but finitely many dyadic rational numbers $\frac{a}{2^b}$, and such that at intervals of differentiability the derivatives are powers of two.

This means that every element of $F$ has the following form. Note that there are only finitely many intervals, each of whose maximum and minimum is a dyadic rational number. The function on each interval has the form $2^ax+\frac{b}{2^c}$, $a, b, c\in\mathbb{Z}$, $c>0$. $$f(x)=\begin{cases} x+\frac12&1\leq x\leq \frac14\\ \frac{x}{2}+\frac18&\frac14\leq x\leq \frac12\\ \frac{x}{4}+\frac{1}{16}&\frac14\leq x\leq \frac12\\ 2x+\frac12&\frac34\leq x\leq \frac78\\ x+1&\frac{29}{32}\leq x\leq 1 \end{cases}$$ One useful way of thinking about $F$ is as a subgroup of automorphisms of the circle, but where we fix a common point. This works because when you pin together the ends of the unit interval you get a circle, and the elements of $F$ fix this pinning point.

The group $T$: The group $T$ is the analogue of $F$ to the circle, so if you additionally allow $F$ to move the pinning point $0=1$. So $T$ is $F$ with rotation. More formally, consider $S^1$ as the unit interval $[0, 1]$, then the group $T$ is the set consisting of piecewise linear homeomorphisms from $S^1$ to itself which are differentiable at all but finitely many dyadic rational numbers $\frac{a}{2^b}$, and such that at intervals of differentiability the derivatives are powers of two.

View $F$ as automorphisms of the circle. Then the group $T$ is simply the group generated by $F$ along with the half-rotation map $x\mapsto x+\frac12\pmod1$, which can be described as follows, $$g(x)=\begin{cases} x+\frac12&1\leq x\leq \frac12\\ x-\frac12&\frac12\leq x\leq 1 \end{cases}$$ Theorem 5.8 and Corollary 5.9 of Cannon, Floyd and Parry's notes combine to prove that $T$ is simple (although this proof uses a different additional generator for $T$ from the one I state, above).

Some more on $F$: These three groups, of which I have defined two, are interesting and much studied. Indeed, I have heard people referred to as "Thompson's group people", and and career's have been built on these groups! Perhaps the most studied of all though if $F$, the only one which is not simple. From the ams article, The group F does not fit easily into standard categories. It is non-Abelian and torsion free and contains a free semigroup on two generators, yet it contains no non-Abelian free subgroup. It is not a matrix group. Every subgroup of F is either finite rank free Abelian or contains an infinite-rank free Abelian subgroup.

One of the current open problems on $F$ is its amenability. See this old MO question and its answers. I have been led to believe that this question is really only interesting because it has garnered so many incorrect answers!

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An example of a group $G$ that has lots of subgroups (15 conjugacy classes of subgroups, 179 total subgroups), but with only the two obvious normal subgroups $1$ and $G$, is the group $G$ of three by three invertible matrices with 0-1 entries and arithmetic operations done mod 2. $G$ has order $(2^3-1)(2^3-2)(2^3-4) = 168$.

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We must look to nonabelian groups since every subgroup of an abelian group will be normal. The example that comes to mind quickly is the alternating group $A_n$ for all $n \geq 5$.

$PSL(3, 2)$ is the second smallest example behind $A_5$ of a group with subgroups, none of whom are normal.

http://groupprops.subwiki.org/wiki/Projective_special_linear_group:PSL(3,2)

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$C_2$ is a perfectly good example of a simple group. In fact, $C_p$ is for every prime $p$. –  Pedro Tamaroff Apr 28 '14 at 3:13
Certainly. I am adding the restriction that nontrivial subgroups actually exist. –  Kaj Hansen Apr 28 '14 at 3:14
Well, you did say "Note that cyclic groups are no good." –  Pedro Tamaroff Apr 28 '14 at 3:15
Ah, I see where the misunderstanding is coming from. The OP had added the restriction that nontrivial subgroups exist. I'll edit my post to clear that up. –  Kaj Hansen Apr 28 '14 at 3:17

A cyclic group of prime order has no subgroups except itself and the group with only one element.

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Hint: Any subgroup of a commutative group is normal. What noncommutative groups do you know.

Bigger Hint: Think about subgroups of $S_5$.

Even Bigger Hint: What about the subgroup of the simple group $A_5$ generated by a product transpositions?

SUPER OMEGA HINT: What is $(12)(45)(12)(34)(12)(45)$?

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