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Let $(V,٭)$ be a group where $V=\{a,b,c,d\}$. $(V,٭)$ has the property that every element is an inverse of itself, so $V$ is called the $V$ group or the "Klein 4 group".

I would like to know whether $(V,٭)$ is isomorphic to $(\mathbb{Z}_4,+_4)$. How may I achieve this? Should I try to show that every element of $(\mathbb{Z}_4,+_4)$ is also an inverse of itself? but I think it may not be!

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You can achieve this by noting that not every element of $(Z,+_4)$ is the inverse of itself, and concluding that the two groups cannot be isomorphic. – Arturo Magidin Dec 5 '11 at 3:51
@ Arturo;thank u,am okay now – neema Dec 5 '11 at 4:14
It's bad for questions to remain unanswered. How about posting a proof that $V$ is not isomorphic to $\mathbb{Z}_4$ as an answer to your own question? – Arturo Magidin Dec 5 '11 at 4:15
@neema You can prove that the only two groups of order 4 up to isomorphism are the Klein 4-group and the cyclic group of order 4. Since $(\mathbb{Z}_4, +_4)$ is isomorphic to the cyclic group of order 4 it cannot be isomorphic to the Klein 4-group. – user38268 Dec 5 '11 at 5:32

Let's answer this question in two ways (largely the same).

Firstly, we might consider the inverses of each element. In the Klein group, every element is its own inverse. In $\mathbb{Z}_4$, neither $1$ ($1 + 1 = 2$) nor $3$ ($3 + 3 = 2$) are their own inverses while $0$ and $2$ are. So they're not isomorphic.

Secondly, we might consider the subgroups of each. What are the subgroups of $\mathbb{Z}_4$? There are only 3 subgroups: the two trivial subgroups and the group $\langle 0, 2 \rangle$. (If $1$ is in the subgroup, then so are $1 + 1 = 2, 1 + 1 + 1= 3$, and $1 + 1 + 1 + 1 = 0$; if 3 is in the subgroup, then so are $3 + 3 = 2, 3 + 3 + 3 = 1, and 3 + 3 + 3 + 3 = 0$). What about the Klein group? There are at least 5 different subgroups: the two trivial groups, and the groups $\langle a,b \rangle$, $\langle a,c\rangle$, and $\langle a,d\rangle$ (where I assumed that $a$ was the identity element). So they're not isomorphic.

Another way equivalent to the second (and let's be honest - to the first as well) is to consider the orders of the elements. In the Klein group, there are 3 elements of order 2 and the identity. In $\mathbb{Z}_4$, there are 2 elements of order 4 ($1$ and $3$), an element of order 2 ($2$), and the identity. So they're not isomorphic.

One should note that the subgroups in the second point are exactly the cyclic subgroups generated by a particular element. Because there are different numbers of elements with the same order, there are different subgroups. One might be made to think that if two groups have the exact same number of elements of each order, than they must be isomorphic. But this is not true! The smallest counterexample are two groups of order 27. And that's pretty cool.

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