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In group theory to prove two groups are isomorphic we have to prove that there exist one-one onto map and also operation is preserved.

For example if we have to disprove that U(10) and U(12) are not isomorphic.

we can use property that $x^2$ =1 $, x \in U(12) $

and take $\phi(9)=\phi(3)\phi(3)=1 $ and $\phi(1)=\phi(1)\phi(1)=1 $

here $\phi(9)$ = $\phi(1)$ $\therefore$ there exist no one-one map.

what if we have to prove or disprove U(x) and U(y) are isomorphic.

Do we have to check all possible numbers.

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No, what you did is enough already. – DonAntonio Mar 25 '13 at 9:56
you mean we always have to find some property to disprove and we cannot prove isomorphism in these type of problems – TLE Mar 25 '13 at 9:58
Well, I didn't mean always. In this particular case what you did is enough, in others it may not be, It all depends. – DonAntonio Mar 25 '13 at 9:59

In general, it can be very difficult to determine whether two given groups are isomorphic, depending on how they are given.

In the specific case of $U(n)$ (I assume by this you mean the group of units module $n$), however, there is a general way.

If we want to compare $U(n)$ and $U(m)$ (I will write it like this here, though it is more commonly referred to as $(\mathbb{Z}/n\mathbb{Z})^{\times}$ or $(\mathbb{Z}/n\mathbb{Z})^*$ since $U(n)$ is also used for other things), then we first write $n = p_1^{a_1}\cdots p_k^{a_k}$ and $m = p_1^{b_1}\cdots p_k^{b_k}$ for distinct primes $p_i$ (in the usual way, we can assume the primes are the same by allowing the exponents to be $0$).

Now, by the Chinese remainder theorem we know that $U(n)\cong U(p_1^{a_1})\times \cdots \times U(p_k^{a_k})$ and similarly for $U(m)$.

If $p$ is an odd prime, $U(p^r)$ is cyclic for any natural number $r$ (and the order is $(p-1)p^{r-1}$).

Finally, for the prime $2$ we have that $U(2)$ and $U(4)$ are cyclic, while for $r\geq 3$ we have $U(2^r) \cong C_2 \times C_{2^{r-2}}$ (I use $C_k$ to mean the cyclic group of order $k$). More precisely, the first factor is generated by $-1$ and the second factor is generated by $5$.

The above allows us to write any $U(n)$ as a product of cyclic groups (where we know the orders), which also allows us to write it as a product of cyclic groups of prime power orders (remember that if $a$ and $b$ are coprime then we have $C_{ab}\cong C_a\times C_b$) and we also know that an abelian group can be written as such in a unique way, so this tells us precisely when the groups will be isomorphic.

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