Theorem 5 (Converse of Lagrange's theorem for abelian groups): Let G be an abelian group of order n and let m be any positive divisor of n. Then there exists a subgroup of G of order m. of order m DEOR FS Before getting to its proof, consider an illustration of this result. You can think of this as a model for giving another proof of this theorem. So, let G = Z₂ XZ₁ ×Z₂×Z¸×Z¸×Z ×Z₁×Z,³ ×Z₁, a group of order 1120210560. of How do we produce a Let us take a divisor of o(G), say 6048. How do we produce a subgroup of G of order 6048? For this, let us first look at the factorisation of 6048, i.e., 6048= 253³7. Then let us find the prime powers of the cyclic group factors of G which just exceed or equal 25,3³ and 7, respectively. Firstly, the exponent 5 of 2 is greater than all the exponents of 2 in o(G). So, we rewrite 5 as 3+2. Similarly, the exponent of 3 is rewritten as 2+1. Then we match them to whatever extent possible, as below. - (Elementary divisors of G) 2³ 2³ 2 3² 3² 3² 5 7³7- | | 2³ 2² 3² 3 7 Now, take a subgroup of the cyclic group in order to match the total exponent that we need for each prime. In this way, we can see that Z₂³×2Z₂³×0×Z¸×3Z₁ ×0×0×0×Z, is a subgroup of the desired order. Note that we have used the fact that every cyclic group has a subgroup of order any number that divides its order. Request explain step-logic of the 27₂3 & 3Z32