Question #5936

the two groups Z / 2 x Z / 2 x Z / 2 x Z/2og Z / 2 ^ 4 both have order 2 ^ 4 = 16 Here denotes as usual Z / m the cyclic group of order m. The argument that these two groups are not isomorphic.

Expert's answer

First notice that each element of G1 has order 2, while G2 has element of order 16.

Indeed, every element x of G1 has the form (a,b,c,d), where a,b,c,d belong to the set {0,1}, and the sum of two

elements

(a,b,c,d) + (a',b',c',d') = (a+a' mod 2,& b+b' mod 2,& c+c' mod 2,& d+d' mod 2).

In particular,

(a,b,c,d) +& (a,b,c,d)& = (2a, 2b, 2c, 2d) = (0,0,0,0) mod 2.

Thus each x from G1 has order 2.

On the other hand, the group Z/16 is cyclic of order 16, and in particular 1 as order 16 in G2.

Now suppose that there exists an isomorphism& q:G2 -> G1.

Then for each n from G2, the element q(n) has the same order as n.

In particular, q(1) has order 16 in G1, which is impossible, since each element of G1 has order 2.

Thus G1 and G2 are not isomorphic.

Indeed, every element x of G1 has the form (a,b,c,d), where a,b,c,d belong to the set {0,1}, and the sum of two

elements

(a,b,c,d) + (a',b',c',d') = (a+a' mod 2,& b+b' mod 2,& c+c' mod 2,& d+d' mod 2).

In particular,

(a,b,c,d) +& (a,b,c,d)& = (2a, 2b, 2c, 2d) = (0,0,0,0) mod 2.

Thus each x from G1 has order 2.

On the other hand, the group Z/16 is cyclic of order 16, and in particular 1 as order 16 in G2.

Now suppose that there exists an isomorphism& q:G2 -> G1.

Then for each n from G2, the element q(n) has the same order as n.

In particular, q(1) has order 16 in G1, which is impossible, since each element of G1 has order 2.

Thus G1 and G2 are not isomorphic.

## Comments

Assignment Expert09.02.12, 15:15Halime07.02.12, 20:34How many different abelian groups, are there of order 16 = 2 ^ 2 (up to isorforfi).

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