The Unapologetic Mathematician

Mathematics for the interested outsider

Split Exact Sequences and Semidirect Products

The direct product of two groups provides a special sort of short exact sequence. We know that there is a surjection G\times H\rightarrow^{\pi_H}H, and we can build an injection G\rightarrow^{{\rm Id}_G\times e_H}G\times H from the identity homomorphism on G and the homomorphism sending everything in G to the identity of H. Since the kernel of the latter is exactly the image of the former, this is an exact sequence {\mathbf1}\rightarrow G\rightarrow^{{\rm Id}_G\times e_H}G\times H\rightarrow^{\pi_H}H\rightarrow{\mathbf1}.

We can do the same thing, swapping G and H, to get a sequence {\mathbf1}\leftarrow G\leftarrow^{\pi_G}G\times H\leftarrow^{e_G\times{\rm Id}_H}H\leftarrow{\mathbf1}. If we compose the homomorphisms H\rightarrow^{e_G\times{\rm Id}_H}G\times H\rightarrow^{\pi_H}H, we get exactly the identity homomorphism on H. When we see this, we say the sequence {\mathbf1}\rightarrow G\rightarrow^{{\rm Id}_G\times e_H}G\times H\rightleftarrows^{\pi_H}_{e_G\times{\rm Id}_H}H\rightarrow{\mathbf1} splits.

Let’s look at this more generally. A split exact sequence is any short exact sequence where the surjection splits: {\mathbf1}\rightarrow N\rightarrow G\rightleftarrows H\rightarrow{\mathbf1}. We can identify N with its image in G, which must be a normal subgroup since it’s also the kernel of the surjection onto H. The homomorphism from H to G must also be an injection — if it had a nontrivial kernel it couldn’t be part of the identity homomorphism from H to itself. We’ll identify H with its image in G as well.

Since N is normal, it’s fixed under conjugation by any element of G. In particular, conjugation by any element of H sends elements of N back into N. This defines a homomorphism \phi:H\rightarrow{\rm Aut}(N). We’ll usually write \phi(h) as \phi_h The action of \phi_h on an element of N is given by \phi_h(n)=hnh^{-1} in the group G. It’s important to remember that we’re considering H and N as living inside the same group G so that this conjugation makes sense.

Okay, so let’s turn this around and build the group G up from the outside. We start with groups N and H, and a homomorphism \phi:H\rightarrow\mathrm{Aut}N. We can take the underlying set of G to be all pairs (n,h) with n in N and h in H. We define composition by (n_1,h_1)(n_2,h_2)=(n_1\phi_{h_1}(n_2),h_1h_2). Verify for yourself that there is an identity and an inverse making this into a group. We call this group the “semidirect product” of N and H, and write N\rtimes_\phi H, or just N\rtimes H if the homomorphism \phi is understood. If \phi_h is the identity automorphism on N for every h, we just have the direct product back.

We can also write down generators and relations like we did for the direct product. If N has generators X and relations R, while H has generators Y and relations S, the semidirect product N\rtimes_\phi H has generators X\cup Y and relations R\cup S\cup\{\phi_h(n)hn^{-1}h^{-1}\}. The elements of N and H don’t commute, but we can “pull h past n” to the right by hitting n with \phi_h: hn=\phi_h(n)h.

One example of a semidirect product I think about is the group of “Euclidean motions” of the plane. We can slide figures around the plane without changing them, and we can turn them around some fixed origin point, but I don’t allow flipping them over. Sliding gives a group T of translations and turning gives a group R of rotations around the origin. Sliding and turning don’t commute: turning a triangle by 90° around the origin and moving it right an inch is different than moving it right an inch and turning it 90° around the origin. However, conjugating a translation by a rotation gives another translation, so the group E of Euclidean motions is the semidirect product T\rtimes R.

Another example that comes up is the wreath product, where H is some subgroup of a permutation group S_n, and N is the direct product of n copies of a group A. We take the action of H on A^n to be permutations of the factors in the product: \phi_h(a_1,...a_n)=(a_{h(1)},...,a_{h(n)}). This one will be useful to us soon.

One last note: as I was thinking about semidirect products in preparation for this post I was trying to determine if there is a universal property for them, like there is for the direct product. I asked a few other people too, and nobody seems to have a good answer. Since I know a few professionals are reading, does anyone know of a universal property that characterizes the semidirect product N\rtimes_\phi H?

March 8, 2007 - Posted by | Algebra, Group theory, Structure of Groups


  1. Actually, Bourbaki (General Topology, Prop. 27) gives such a universal property: Let $f \colon N \to G$, $g \colon H \to G$ be two homomorphisms into a group $G$, such that \[f(\phi_h(n)) = g(h)f(n)g(h^{-1})\] for all $n \in N$, $h \in H$. Then there is a unique homomorphism $k colon N \rtimes H \to G$ extending $f$ and $g$ in the usual sense.

    Comment by Dvir | September 16, 2007 | Reply

  2. Just a small typo: when you define the semidirect product, you should have written “a homomorphism \phi:H -> Aut(N)” and, in the following line, \phi_{h_1} instead of \phi_h.
    I hope not to say stupid things, and, by the way, thank you for all these free lessons!

    Comment by edriv | December 10, 2007 | Reply

  3. thanks. tweaked.

    Comment by John Armstrong | December 10, 2007 | Reply

  4. […] Weyl group of is then the subgroup of the wreath product consisting of those transformations with an even number of flips coming from the components. […]

    Pingback by Construction of D-Series Root Systems « The Unapologetic Mathematician | March 3, 2010 | Reply

  5. […] don’t have any restrictions on how many signs we can flip, the Weyl group for is exactly the wreath product […]

    Pingback by Construction of B- and C-Series Root Systems « The Unapologetic Mathematician | March 4, 2010 | Reply

  6. […] On the other hand, given an arbitrary automorphism , it sends to some other base . We can find a sending back to . And so ; it’s an automorphism sending to itself. That is, ; any automorphism can be written (not necessarily uniquely) as the composition of one from and one from . Therefore we can write the automorphism group as the semidirect product: […]

    Pingback by The Automorphism Group of a Root System « The Unapologetic Mathematician | March 11, 2010 | Reply

  7. Typo: In the second paragraph, second longer formula, you have \pi_G taking values in H.

    Comment by Tommi Brander | January 11, 2013 | Reply

  8. Thanks; fixed.

    Comment by John Armstrong | January 11, 2013 | Reply

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