I need to show that $(\mathbb{Z} / 12 \mathbb{Z})^{*} \cong \mathbb{Z} / 2 \mathbb{Z} \times \mathbb{Z} / 2 \mathbb{Z}$.

But I don't even know where to begin. Is there a smart way to show this where I don't have to come up with an actual isomorphism between the two groups? Showing that two groups are isomorphic is one of the hardest problems that I have encountered since it's often difficult to invent an isomorphism between the two groups.

I have encountered a similar exercise where I had to show
$(\mathbb{Z} / 14 \mathbb{Z})^{*}\cong(\mathbb{Z} / 2 \mathbb{Z}) \times(\mathbb{Z} / 3 \mathbb{Z})$
But this was much simpler since I showed that $|(\mathbb{Z} / 14 \mathbb{Z})^{*}|=\varphi(14)=6$ and that $(\mathbb{Z} / 14 \mathbb{Z})^{*}$ is a cyclic group.
So I had a theorem that said that a cyclic group of order 6 is isomorphic to the quotient group $\mathbb{Z} /6 \mathbb{Z}$.
Furthermore, $6=2\cdot3$ and $gcd(2,3)=1$ so the CRT tells me that $\mathbb{Z} /6 \mathbb{Z}\cong \mathbb{Z} / 2 \mathbb{Z} \times \mathbb{Z} / 3 \mathbb{Z}$.
That is $(\mathbb{Z} / 14 \mathbb{Z})^{*}\cong\mathbb{Z} /6 \mathbb{Z}\cong\mathbb{Z} / 2 \mathbb{Z} \times \mathbb{Z} / 3 \mathbb{Z}$

However, none of these methods work in the case above since $gcd(2,2)\neq1$ and the group $(\mathbb{Z} / 12 \mathbb{Z})^{*}$ is not cyclic.

  • 3
    $\begingroup$ More generally, $(\mathbb Z/{mn}\mathbb Z)^*\cong \mathbb (Z/m\mathbb Z)^*\times(\mathbb Z/n\mathbb Z)^*$ when $\gcd(m,n)=1.$ $\endgroup$ May 18, 2022 at 23:29
  • $\begingroup$ I mean you could just do it by figuring out the units of $\mathbb Z/12\mathbb Z$ and working out the multiplication table, finding that $u^2=1$ for all elements. There aren’t many groups of $4$ elements. $\endgroup$ May 18, 2022 at 23:36
  • $\begingroup$ @ThomasAndrews I have already checked that all elements are of order 2. But I don't quite see how that helps me. $(\mathbb{Z} / m n \mathbb{Z})^{*} \cong(Z / m \mathbb{Z})^{*} \times(\mathbb{Z} / n \mathbb{Z})^{*}$ when $\operatorname{gcd}(m, n)=1$ is not in my book, so I would have to prove that it is true for the group of units. $\endgroup$
    – hh25
    May 18, 2022 at 23:52
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    $\begingroup$ In general, finding an isomorphism between two given groups can be hard, but in this particular case it's easy. Any bijection that sends the identity to the identity works. $\endgroup$ May 18, 2022 at 23:52

2 Answers 2


The most obvious way to show this is by using the Chinese Remainder Theorem to see that ${\mathbb Z}_{12} \cong {\mathbb Z}_4 \times {\mathbb Z}_3$ (as rings) and therefore also ${\mathbb Z}_{12}^* \cong {\mathbb Z}_4^* \times {\mathbb Z}_3^*$ (as multiplicative groups). Now ${\mathbb Z}_4^*$ and ${\mathbb Z}_3^*$ both have two elements, so both are cyclic of order $2$.

This approach generalizes to finding ${\mathbb Z}_m^*$ and reduces it to finding ${\mathbb Z}_{p^k}^*$ for a prime $p$. (Which is a cyclic group of order $(p-1)p^{k-1}$, except if $p = 2$. There ${\mathbb Z}_2^* \cong \{1\}$, ${\mathbb Z}_4^* \cong {\mathbb Z}_2$, and ${\mathbb Z}_{2^k}^* \cong {\mathbb Z}_2 \times {\mathbb Z}_{2^{k-2}}$ for $k \geq 3$).

If for some reason this is not an acceptable argument, then the answer of course depends on what acceptable arguments are.

A more primitive way of seeing that ${\mathbb Z}_{12}^* \cong {\mathbb Z}_2 \times {\mathbb Z}_2$ is noting that the elements of ${\mathbb Z}_{12}^*$ are $\overline{1}$, $\overline{5}$, $\overline{7}$, and $\overline{11}$. Then work out the multiplication table and see that it is the same, after renaming, as that of ${\mathbb Z}_2 \times {\mathbb Z}_2$.

  • $\begingroup$ Thank you for your reply. One thing I don't quite understand is the implication $\mathbb{Z}_{12} \cong \mathbb{Z}_{4} \times \mathbb{Z}_{3}$ (as rings) $\implies$ $\mathbb{Z}_{12}^{*} \cong \mathbb{Z}_{4}^{*} \times \mathbb{Z}_{3}^{*}$ (as multiplicative groups). Is it possible you can elaborate on that? $\endgroup$
    – hh25
    May 20, 2022 at 11:14
  • $\begingroup$ This is just in general: $(R \times S)^* = R^* \times S^*$ (not even just $\cong$), i.e., $(r,s)$ is a unit in $R \times S$ if and only if $r$ is a unit in $R$ and $s$ is a unit in $S$. $\endgroup$ May 20, 2022 at 12:27

Do you know what $H=(\mathbb{Z}/12\mathbb{Z})^*$ looks like as a set?

You'll notice it has four elements and since it's a subgroup of an abelian group, it's abelian. That only leaves a couple of possibilities:

If $H$ has an element of order $4$ then it's isomorphic to $\mathbb{Z}/4\mathbb{Z}$, and if it doesn't it's isomorphic to $(\mathbb{Z}/2\mathbb{Z})^2$ (why?). So at this point you can just check the order of the elements manually.

  • $\begingroup$ I think there is something I have misunderstood or forgotten. It's not obvious to me how the order of H and that it's abelian group imply isomorphism to the two groups. Can you maybe elaborate? $\endgroup$
    – hh25
    May 18, 2022 at 23:43
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    $\begingroup$ $H$ is an abelian group of exactly $4$ elements. There are only $2$ such groups, up to isomorphism. Now $4$ is a small enough number so that you can verify this yourself. To see: Let $H$ be a group with $4$ elements and then let $\alpha \not = e$ be an element in $H$. If $\alpha^2 = e$ then there must be another element $\beta$. Then $\beta^2$ must be $e$ as well, and then the one remaining element must be $\alpha\beta$. Compare this to $\mathbb{Z}/2 \times \mathbb{Z}/2$. If $\alpha^2 \not = e$ then as $H$ has $4$ elements $H=\{\alpha,\alpha^2,\alpha^3,e\}$. $\endgroup$
    – Mike
    May 18, 2022 at 23:58
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    $\begingroup$ @Mike Okay, I think I understand your reasoning. So if G is an abelian group and $|G|=8$, then you would say it could be isomorphic to $H=(\mathbb{Z} / 8 \mathbb{Z})$ or $K=(\mathbb{Z} / 2 \mathbb{Z})\times (\mathbb{Z} / 4 \mathbb{Z})$ or $(N=\mathbb{Z} / 2 \mathbb{Z})^3$ since these groups are all abelian groups with 8 elements. H has a group element of order 8, K has a group element of maximum lcm(2,4)=4, and N has a group element of maximum order lcm(2,2,2)=2. So by checking the order of the group elements of G I could determine which of the groups it's isomorphic to. $\endgroup$
    – hh25
    May 19, 2022 at 1:46
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    $\begingroup$ But I think I understand the problem of confusion. The above reasoning presumes the fact that an abelian group of order n has to be isomorphic to the abelian group $\mathbb{Z} / n \mathbb{Z}$ or any of the product groups with the same order. However, this is not obvious to me and not stated in my book. So it's something I would have to prove anyway. So I guess there is only one way of showing this for me, which is by creating a map between the two groups and show it is an bijective group homomorphism. $\endgroup$
    – hh25
    May 19, 2022 at 1:52
  • $\begingroup$ One way you could go about arguing that there are only two abelian groups of order four (up to isomorphism) would be to split into cases. If there's an element of order $4$ then you immediately know it's $\mathbb{Z}/4\mathbb{Z}$. Otherwise all (non-identity) elements have order $2$. But then you can construct an isomorphism between this group and $(\mathbb{Z}/2\mathbb{Z})^2$. There's also a big theorem about classifying such abelian groups, but if you haven't seen this then it's likely unfair to use $\endgroup$
    – Theo C.
    May 19, 2022 at 3:23

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