Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $G$ be a finite group and let $N \unlhd G$ be a normal subgroup of $G$. I would like to embed $G$ in $(G/N) \times A$ for some small group $A$. I require the embedding to map $g \in G$ to $(gN, a_g)$ for some $a_g \in A$. What is the smallest $A$ I can take?

share|cite|improve this question
Universally, $A=G$. – user641 May 18 '11 at 22:11
Do you want the embedding to be a homomorphism? – Grumpy Parsnip May 18 '11 at 22:54
up vote 8 down vote accepted

You can always take A = G, and often you cannot do better. If G is cyclic of order 4 and N is the normal subgroup of index 2, then the smallest A that works is of course G, since G/N × A must at the very least have an element of order 4.

Another silly case:

  • When N = 1, you can always take A = 1.
  • When N = G, the minimal choice is A = G.

Consider the homomorphism φ from G to G/N × A given by g → ( gN, f(g) ). Notice that f must be a homomorphism from G to A. What is the kernel of φ? It is just the intersection of the kernel of f with N. Hence we take A to be minimal amongst all G/M where MN = 1.

For instance, if G is a p-group with a cyclic center and N ≠ 1, then A = G is minimal. If G is dihedral of order 2p and N ≠ 1, then A = G is minimal. If G is dihedral of order 2pq and N has order p, then one can take A to be dihedral of order 2p, that is M has order q. Here pq are distinct odd primes.

In particular, if N contains the socle of G, then A = G is minimal. The converse also holds, since otherwise M being a minimal normal subgroup not contained in N will work for A = G/M.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.