# Group Theory Isomorphism $|G|=10$ and $\mathbb{Z}_{10}$

I think I just didn't get the core of group theory. Although it makes sense to me to follow the regular steps to solve problems of group theory. For example, a group of order $10$ is isomorphic to $\mathbb{Z}_{10}$. To prove this, the standard solution suggests that we have to suppose that there are $2$ elements $x,y$. $x$ has order of 5 another has order of $2$ to begin with. And finally applied that it's isomorphic to $\mathbb{Z}_{2}\times\mathbb{Z}_{5}$ and then isomorphic to $\mathbb{Z}_{10}$

I am confused that, isn't that obvious that a group of order 10 is isomorphic to Z10?? Both of them have $10$ elements. We can simply project them one-by-one.. like $1$ to $x$ ; $2$ to $e$ ; $3$ to $y$.... ....... .......

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It is not true that a group of order 10 must be cyclic: the group of rigid motions of the regular pentagon has order $10$ and is not abelian. –  Arturo Magidin Mar 27 '12 at 21:37

The idea behind an isomorphism isn't just that the groups have the same number of elements, but also that they work the same. You can surely make a bijective mapping $\phi:G\rightarrow \mathbb{Z}_{10}$ for any group $G$ of order $10$, but to be isomorphic, you must have that $\phi$ is a homomorphism. Recall the definition: $\phi:G\rightarrow H$ is a homomorphism if $\phi(xy)=\phi(x)\phi(y)$ for all $x,y\in G$. Think for a moment about the intuitive meaning of that statement.

• $\phi$ is a mapping which associates elements in $G$ to elements in $H$.
• $xy\in G$is the product of two elements in $G$; the resulting element is determined by the algebraic structure of $G$, the way its elements interact with one another.
• $\phi(x)\phi(y)$ is the product of the elements associated with $x$ and $y$ under the mapping $\phi$. The product takes place in $H$, so the resulting element depends on the algebraic structure of $H$.
• If $\phi$ is bijective, and if $\phi(xy)=\phi(x)\phi(y)$ no matter which $x$ and $y$ you pick, then given any product $xy$ in $G$, the corresponding product in $H$ works the same as the one in $G$. Thus the groups work the same, so they are the same in the eyes of a group theorist.

If such a mapping does not exist, then at least some part of the structures of $G$ and $H$ is irreconcilably different, so the groups themselves are different.

In the example of $\mathbb{Z}_{10}$ vs $D_{10}$, we can prove that no isomorphism exists by noticing that $D_{10}$ has no element of order $10$, while $Z_{10}$ does - an essential difference in structure which would have to be preserved by an isomorphism. (It is easy to prove that isomorphisms preserve element orders; you should try it.)

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It seems like you're confused as to what an isomorphism actually is. Surely two sets of order 10 are isomorphic as sets but as the other answers have said this doesn't necessarily make them isomorphic as groups. For two groups to have a (group) isomorphism you need a bijective (group) homomorphism to exist between them. This is why your projection idea wont work (it isn't enough for a group isomorphism), try it on the dihedral group of degree 5 and the cyclic group of order 10, none of the bijections will be homomorphisms.

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Thanks!I forget the definition that every "cyclic group of order n" is isomorphic to Z10. And of course, it can't always guarantee the homomorphism –  Jinji Mar 28 '12 at 1:14
@Jinji: It is false that "every cyclic group of order n" is isomorphic to $Z_{10}$. Every cyclic group of order 10 is isomorphic to $Z_{10}$, but not every cyclic group of order $n$. –  Arturo Magidin Mar 28 '12 at 4:45

It is false that a group of order $10$ must be cyclic.

There are two isomorphism types of groups of order $10: an abelian group, which is indeed cyclic, and a nonabelian group. The nonabelian group of order$10$is the dihedral group of degree$5$(you may see it denoted as either$D_5$or$D_{10}$). It can be realized as the group of rigid motions of a regular pentagon. It has presentation: $$\Bigl\langle r,s\;\Bigm|\; r^5 = s^2 = 1, sr=r^{4}s\Bigr\rangle.$$ But with your final paragraph: it is not enough for them to have the same number of elements. For example,$\mathbf{Z}_2\times\mathbf{Z}_2$and$\mathbf{Z}_4$both have 4 elements, are both abelian, but they are not isomorphic, because the latter group has only two solutions to$x+x=0$(namely,$x=0$and$x=2$), but the former group has four solutions to that equation (every element is a solution). So you don't know ahead of time that every group of order$10$(or even that every abelian group of order$10\$) must be cyclic.

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