[Math] how to show that the groups are not isomorphic

abstract-algebragroup-isomorphism

$Z_2\times Z_2\times Z_3$ , $Z_4 \times Z_3$ , $D_{12}$, $A_4$

Show that no 2 groups are isomorphic to each other?

$Z_2 \times Z_2 \times Z_3$ and $Z_4 \times Z_3$): For $Z_2 \times Z_2 \times Z_3$ and $Z_4 \times Z_3$ I said that they are not isomorphic because $Z_2 \times Z_2 \times Z_3$ has an LCM of 6 vs. 12 so they aren't isomorphic. Am I right?

$Z_2 \times Z_2 \times Z_3$ and $D_{12}$: I am not sure, but I think because the order of $Z_2 \times Z_2 \times Z_3$ is 6 and $D_{12}$ is 12, they are not isomorphic?

$Z_2 \times Z_2 \times Z_3$ and $A_4$: ? Not sure?

$Z_4 \times Z_3$ and $D_{12}$: ?

$Z_4 \times Z_3$ and $A_4$: ?

$D_{12}$ and $A_4$: I said they aren't isomorphic because $D_{12}$ has a rotation of order 12 and $A_4$ has order 1, 2 or 3 so they are not. ( Is this correct?)

I'd appreciate if you can tell me if I am right / wrong for the ones I solved and help me solve the remaining ones. Thanks!

Best Answer

The first two groups are abelian but are not isomorphic because they have different exponents. Actually, the second group is cyclic but the first one is not.

The last two groups not abelian and so cannot be isomorphic to the any of the first two groups.

$D_{12}$ has an element of order $6$ but $A_4$ doesn't and so they are not isomorphic.

Related Solutions

[Math] Why is $\Bbb Z_2 \times \Bbb Z_4$ not isomorphic to $\Bbb Z_8$

Indeed, they do have a one-to-one correspondence between, them (in fact, there are $8!$ such correspondences). The problem is that they don't have the "same structure," so none of these correspondences will be homomorphisms, and so no isomorphism exists. But it's tedious in the extreme to check so many functions to see if any of them is a homomorphism, so we instead make use of one of many properties preserved under isomorphisms. Two such (related) properties are that:

isomorphisms preserve orders of elements (meaning that any given element will be mapped to an element of the same order), and
isomorphic images of cyclic groups are again cyclic groups.

Those two properties are probably the simplest way to show that these two groups are not isomorphic, since (in particular) there are four different order $8$ elements of $\Bbb Z_8,$ but no such elements in $\Bbb Z_2\times\Bbb Z_4.$ So, we can directly conclude that they are not isomorphic by fact 1. Alternately, we can note that this means $\Bbb Z_8$ is cyclic and $\Bbb Z_2\times\Bbb Z_4$ is not, so can indirectly conclude that they are not isomorphic by fact 2.

There are many other such properties that we can use in other cases. It's good to keep several of them in mind. As Mark points out in the comments, though, the first property is a necessary condition for a function to be an isomorphism, but not sufficient. However, two cyclic groups of the same order will necessarily be isomorphic. Unfortunately, most groups aren't cyclic, so this property won't usually be useful.

[Math] Find the smallest positive integer n such that there are exactly four non-isomorphic abelian groups of order n

The number of non-isomorphic Abelian groups depends on the partition number of the exponents in the prime factorization. Let $P(k)$ be the number of ways to partition $k$ (for example, $P(4) = 5$, since we have $4, 3+1, 2+2, 2+1+1$, and $1+1+1+1$). Then for $n=p_1^{e_1} \ldots p_r^{e_r}$, there are $$\prod_i P(e_i)$$ non-isomorphic Abelian groups of order $n$. You missed $2+2$ among the ways to partition $4$, corresponding to $\mathbb Z/(2^2\mathbb Z) \times \mathbb Z/(2^2\mathbb Z) = \mathbb Z/4\mathbb Z \times \mathbb Z/4\mathbb Z$.

You can use this formula to answer the original question. The first few partition numbers are:

P(1) = 1    1
P(2) = 2    2, 1+1
P(3) = 3    3, 2+1, 1+1+1
P(4) = 5    4, 3+1, 2+2, 2+1+1, 1+1+1+1
P(5) = 7    5, 4+1, 3+2, 3+1+1, 2+2+1, 2+1+1+1, 1+1+1+1+1

and they obviously increase from here. Now, if there are $4$ Abelian groups of order $n$, then we must have $e_1=e_2=2$, and we choose as bases the two smallest primes. Thus the number we are looking for is $2^23^2=36$.

Best Answer

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[Math] Why is $\Bbb Z_2 \times \Bbb Z_4$ not isomorphic to $\Bbb Z_8$

[Math] Find the smallest positive integer n such that there are exactly four non-isomorphic abelian groups of order n

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