Join of subnormal subgroups is subnormal iff their commutator is subnormal

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Statement

Suppose H,K are Subnormal subgroup (?)s of a group G. Then, the following are equivalent:

  1. The join of subgroups \langle H, K \rangle is a subnormal subgroup of G.
  2. The subgroup H^K, i.e., the closure of H under the action of K by conjugation, is a subnormal subgroup of G.
  3. The commutator of two subgroups [H,K] is a subnormal subgroup of G.

Moreover, if s_1, s_2, s_3 denote the subnormal depths of the subgroups in (1),(2),(3) respectively, and h,k denote the subnormal depths of H,K, we have:

s_2 \le s_1 + 1, s_3 \le s_2 + 1, s_3 \le s_1 + 1, s_2 \le s_3h, s_1 \le s_2k.

Related facts

Facts used

  1. Commutator of two subgroups is normal in join: For H, K \le G, [H,K] is normal in \langle H, K\rangle. Further, we have H[H,K] = H^K, K[H,K] = H^K, and KH^K = \langle H, K \rangle.
  2. Subnormality is normalizing join-closed: For A, B \le G subnormal of depths a,b, with B \le N_G(A), we have that AB is subnormal of depth at most ab.

Proof

Given: A group G with subnormal subgroups H, K.

To prove: Consider the subgroups \langle H, K \rangle, H^K,, and [H,K]. If any one of them is subnormal, so are the others. Further, if s_1, s_2, s_3 are their subnormal depths, we have:

s_2 \le s_1 + 1, s_3 \le s_2 + 1, s_3 \le s_1 + 1, s_2 \le s_3h, s_1 \le s_2k.

Bounding the commutator in terms of the join

To prove: If \langle H, K \rangle is subnormal of depth s_1 , so is [H,K], and if s_3 denotes its subnormal depth, we have s_3 \le s_1 + 1.

Proof: By fact (1), the commutator [H,K] is a normal subgroup of \langle H, K \rangle, and thus, if \langle H, K \rangle is subnormal of depth s_1, we have a subnormal series for [H,K] of length s_1 + 1. This yields s_3 \le s_1 + 1.

Bounding the commutator in terms of the closure

To prove: If H^K is subnormal of depth s_2, then [H,K] is subnormal of depth s_3 with s_3 \le s_2 + 1.

Proof: By fact (1), [H,K] is normal in \langle H, K \rangle, and in particular, is normal in the intermediate subgroup H^K. Thus, if H^K has subnormal depth s_2, we obtain a subnormal series for [H,K] of length s_2 + 1, yielding s_3 \le s_2 + 1.

Bounding the closure in terms of the join

To prove: If \langle H, K \rangle is subnormal of depth s_1, then H^K is subnormal of depth s_2 with s_2 \le s_1 + 1.

Proof: We have KH^K = \langle H, K \rangle, and by definition, K normalizes H^K. Thus, H^K is normal in \langle H, K \rangle. If \langle H, K \rangle is subnormal of depth s_1, we get a subnormal series for H^K of length s_1 + 1, yielding H^K is subnormal with subnormal depth at most s_1 + 1.

Bounding the closure in terms of the commutator

To prove: If [H,K] is subnormal of depth s_3 and H is subnormal of depth h, then H^K is subnormal of depth at most s_3h.

Proof: By fact (1), H normalizes [H,K] and H^K = H[H,K]. Thus, by fact (2), H^K is subnormal of depth at most s_3h.

Bounding the join in terms of the closure

To prove: If H^K is subnormal of depth s_2 and K is subnormal of depth k, then \langle H, K \rangle is subnormal of depth at most s_2k.

Proof: We have KH^K = \langle H, K \rangle, with K normalizing H^K. Thus, by fact (2), \langle H, K \rangle is subnormal of depth at most s_2k.

References

Textbook references

  • A Course in the Theory of Groups by Derek J. S. Robinson, ISBN 0387944613, More info, Page 388, Theorem 13.1.6, Section 13.1 (Joins and intersections of subnormal subgroups)