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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?

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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
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1 Answer 1

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.

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