# Examples of finite groups that are not a semidirect product

I'm looking for examples of (families of) finite groups that are not semidirect products.

When first learning group theory, the first such group that one encounters is $Q_8$. In my search for other groups that are not semidirect products, the only examples I could find were simple groups, which clearly cannot be semidirect products since they don't have normal subgroups.

Does anyone have example of non-simple finite groups that are not semidirect products?

• All the generalized quaternion groups satisfy this, as they have a unique element of order $2$. – Tobias Kildetoft Jul 24 '13 at 18:04
• Cyclic groups of prime-power order also work (or, as it were, don't work - they are not semidirect products). Note that these are not always simple, 'cause I said prime-power. – user1729 Jul 24 '13 at 18:11
• quasi-simple groups (nonsplit central extensions of simple groups) also work. For instance SL(n,q) or the valentiner group. – Jack Schmidt Jul 24 '13 at 18:49

The easiest examples of non-simple groups, that are not a semidirect product of two non-trivial subgroups, are cyclic groups of order $$p^n$$, where $$p$$ is prime and $$n \geq 2$$, and generalized quaternion groups

The name for what you are looking for is a "splitting-simple group".

If you are looking for the smallest non-nilpotent example, that would be a group of order 48, the tilde covering of $$S_4$$. It's actually pretty easy to see that it's the smallest non-nilpotent example. Due to the existence of Hall subgroups, a solvable group that is not a $$p$$-group will be a semidirect product if it has a normal Sylow subgroup. Using Sylow's theorems and the property of the Fitting subgroup mentioned below, you can find that there are three such groups of order less than 48: $$S_4$$ and its two covering. However, $$S_4$$ is a semidirect product in another way ($$2^2$$ by $$S_3$$) as is its other double cover (GL(2,3), which is $$Q$$ by $$S_3$$). However, the tilde cover has a unique element of order 2, making it clear it can't be a semidirect product of two groups of even order, and it can't by odd by even or even by odd, either, since the Sylows are not normal and 3*8 is the only way to write 24 non-trivially as even*odd.

The best description of the tilde cover of $$S_4$$ is here, showing its faithful 2-dimensional representation over GF(9).

With the possible exceptions of orders 32 and 64, the only splitting simple groups of order less than 81 are $$A_5$$, the tilde cover of $$S_4$$, the generalized quaternion (including quaternion) groups, and the cyclic groups of prime power order.

To see how the reasoning works: If a splitting-simple group of order 72 exists, its Fitting subgroup must be non-cyclic abelian of order 12, $$O_3$$ must be central, and the quotient by $$O_3$$ must be $$S_4$$. $$O_3$$ cannot then be in the derived group, since $$S_4$$ doesn't have a triple cover. But then the derived group has order 12 (it must contain the pre-image of $$A_4$$ without meeting $$O_3$$). Thus, all elements of order 2 must be in $$O_2$$ lest we have a semidirect product of $$A_4$$ by 6. But since the derived group has index 6, it is contained in an index 3 normal subgroup that doesn't meet $$O_3$$--thus the group is a semidirect product after all.

Also, note that not all non-solvable examples are simple. A cover of a simple group will be splitting-simple, so, for example, at order 120, although $$S_5$$ is not splitting-simple, $$SL(2,5)$$, the universal cover of $$A_5$$, is splitting-simple. In fact it is not hard to see that $$SL(2,5)$$ is the only splitting simple group of order 120, as you can rule out solvable examples using the theorem that the Fitting subgroup of a solvable group contains it's own centralizer.