While searching for non-isomorph subgroups of order 2002 I just encountered something, which I want to understand. Obviously I looked for abelian subgroups first and found 2002=2^2 cdot 503 so we have the groups Z/2^2 Z times Z/503Z, Z/2Z times Z/2Z times Z/503Z

Liberty Page 2022-09-20 Answered
Nonisomorph groups of order 2002
While searching for non-isomorph subgroups of order 2002 I just encountered something, which I want to understand. Obviously I looked for abelian subgroups first and found 2002 = 2 2 503 so we have the groups
Z / 2 2 Z × Z / 503 Z , Z / 2 Z × Z / 2 Z × Z / 503 Z
Now I want to understand why those two are not isomorph. I know that for two groups Z / n Z × Z / m Z Z / ( n m ) Z it has to hold that gcd ( n , m ) = 1. But I don't understand how we can compare Groups written as two products with groups written as three products as above, how does that work? And I think that goes in the same direction: How is it then at the same time that
Z / 4 Z × Z / 503 Z Z / 2012 Z Z / 2 Z × Z / 2 Z × Z / 503 Z Z / 2 Z × Z / 1006 Z
because gcd ( 4 , 2012 ) 1 , gcd ( 2 , 2 ) 1 , gcd ( 503 , 1006 ) 1. I don't understand the difference to the first comparison.
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Answers (2)

espovilham7
Answered 2022-09-21 Author has 10 answers
Step 1
Here's where the hypothesis gcd ( n , m ) = 1 plays a role : if d := gcd ( n , m ) 1, then Z / n m Z has an element of order d 2 but Z / n Z × Z / m Z does not.
Step 2
In your example, Z / 4 Z × Z / 503 Z has an element of order 4 (namely (1,0)), but Z / 2 Z × Z / 1006 Z has no element of order 4.

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Celinamg8
Answered 2022-09-22 Author has 1 answers
Step 1
First let's note that 2 2 503 = 2012 2002
Abelian groups of order 2002:
There is only one Abelian group of order 2002, namely
Z 2002 Z 2 × Z 7 × Z 11 × Z 13
Abelian groups of order 2012:
Since 2012 = 2 2 503 it might initially seem like there are four possibilities:
Z 2 × Z 2 × Z 503
Z 4 × Z 503
Z 2 × Z 1006
Z 2012
Step 2
But using the fact that
Z n × Z m Z n m iff G C D ( n , m ) = 1.
we get that
Z 2 × Z 2 × Z 503 Z 2 × Z 1006
and
Z 4 × Z 503 Z 2012
and
Z 2 × Z 1006 Z 2012
Hence there are exactly two non-isomorphic groups of order 2012.

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