Join of normal and subnormal implies subnormal of same depth

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This article describes a computation relating the result of the Join operator (?) on two known subgroup properties (i.e., Normal subgroup (?) and Subnormal subgroup (?)), to another known subgroup property (i.e., Subnormal subgroup (?))
View a complete list of join computations


Let k \ge 1 be a natural number. Then, the join of any normal subgroup of a group with any k-subnormal subgroup is k-subnormal (Note: When we say a subgroup is k-subnormal, we mean that its subnormal depth is at most equal to k).

In symbols, if K is a k-subnormal subgroup of G and H is a normal subgroup of G, then the join of subgroups \langle H, K \rangle (which in this case is also the product of subgroups HK) is also a k-subnormal subgroups.

Facts used

  1. Join lemma for normal subgroup of subgroup with normal subgroup of whole group: This states that if A \underline{\triangleleft} B \le G and C \underline{\triangleleft} G, then \langle A, C \rangle \underline{\triangleleft} \langle B, C \rangle.


Given: A group G, a normal subgroup H of G, a k-subnormal subgroup K of G.

To prove: \langle H, K \rangle is k-subnormal in G.


Step no. Assertion Given data used Facts used Previous steps used Explanation
1 There exists a subnormal series from K to G given as follows: \! K = K_0 \underline{\triangleleft} K_1 \underline{\triangleleft} \dots \underline{\triangleleft} K_n = G. Here, \underline{\triangleleft} means is normal in K is k-subnormal in G Using the definition of k-subnormal.
2 For 0 \le i \le n - 1, \langle H, K_i \rangle \underline{\triangleleft} \langle H, K_{i + 1} \rangle. H is normal in G. (1) (1) For fact (1), set A = K_i, B = K_{i+1}, and C = H.
3 The series \langle H, K \rangle = \langle H, K_0 \rangle \underline{\triangleleft} \langle H, K_1 \rangle \underline{\triangleleft} \dots \underline{\triangleleft} \langle H, K_n \rangle = G is a subnormal series of length k for \langle H, K \rangle in G. (2) Follows directly.
4 \langle H, K \rangle is k-subnormal in G. (3) Follows directly.