For two groups $G_1, G_2$ and two central subgroups $U_1 \le Z(G_1), U_2 \le Z(G_2)$ which are isomorphic by some given $\mu : U_1 \to U_2$ the central product is the group $$ (G_1 \times G_2) / D $$ with $D = \{ (u_1, \mu(u_1)^{-1}) : u_1 \in U_1 \}$, i.e. we identify $U_1$ and $U_2$ in the direct product. Obviously for abelian groups every subgroup is central. So now my question concerns this product and cyclic groups.

Let $G_i = \langle g_i \rangle$ ($i = 1,2$) be cyclic groups of order $p^{a_i}$ with $a_1 \ge a_2 > n$. Set $$ U_i = \langle g_i^{p^{a_i - n}} \rangle $$ the subgroup of order $p^n$ of $G_i$. Let $\mu$ be the isomorphism of $U_1$ onto $U_2$ with $$ \mu( g_1^{p^{a_1 - n}} ) = g_2^{p^{a_2 - n}}. $$

Then the central product of $G_1$ and $G_2$ with $U_1$ and $U_2$ identified by $\mu$ is isomorphic to the direct product of cyclic groups of order $p^{a_1}$ and $p^{a_2 - n}$.

I have some arguments, but I do not find them satisfying. But I post what I have done.

My approach: I switched notation and write $\mathbb Z_n$ for the cyclic group of order $n$, and denote them additively with elements $\mathbb Z_n = \{0,1,\ldots, n-1\}$. Then for $G_1 = \mathbb Z_{p^{a_1}}$ and $G_2 = \mathbb Z_{p^{a_2}}$ we have $$ U_1 = \{ 0, p^{a_1 - n}, 2\cdot p^{a_1 - n}, 3\cdot p^{a_1 - n}, \ldots, (p^n -1) \cdot p^{a_1 - n} \} \cong \mathbb Z_{p^n} $$ and similar for $U_2$. Now we have $$ D = \{ (k \cdot p^{a_1 - n}, k \cdot p^{a_2 - n}) : k = 0,1,\ldots, n-1 \} $$ and a coset has the form $$ (x,y) + D = \{ (x + k \cdot p^{a_1 - n}, y + k \cdot p^{a_2 - n}) : k = 0,1,\ldots, n - 1 \}. $$ Then we have $p^{a_1} \cdot p^{a_2 - n}$ different cosets which are $$ \begin{array}{cccc} (0,0) + D & (1,0) + D & \ldots & (p^{a_2} - 1, 0) + D \\ (0,1) + D & (1,1) + D & \ldots & (p^{a_2} - 1, 1) + D \\ \vdots & \vdots & & \vdots \\ (0,p^{a_2 - n} - 1) + D & (1, p^{a_2 - n}-1) + D & \ldots & (p^{a_2} - 1, p^{a_2 - n} - 1) \end{array} $$ First every $(x,y)$ is contained in at least one of the above listed cosets, for write $y = k \cdot p^{a_2-n} + r$ with $0 \le r < p^{a_2-n}$, then $(x,y) + D = (x - k\cdot p^{a_1}, r) + D$ with $x - k\cdot p^{a_1} \in \mathbb Z_{p^{a_1}}$. Also all of the listed cosets are different, for if $(x, r) + D, (\hat x, \hat r) + D$ are two with $0 \le x,\hat x < p^{a_2}$ and $0 \le r,\hat r < p^{a_1 - n}$ and $r > \hat r$, then $$ (x,r) - (\hat x, \hat r) = (x - \hat x, r - \hat r) $$ and $0 < r - \hat r < p^{a_2 - n}$ and hence $r - \hat r$ is not divisble by $p^{a_2 - n}$, which shows that $(x - \hat x, r - \hat r) \notin D$. So as the addition is componentwise on the cosets, we can "see" that they form a group isomorphic to $\mathbb Z_{p^{a_1}} \times \mathbb Z_{p^{a_2 - n}}$.

That is the best I can write down, but I am not glad with it. I think it is to much computation and "messy". Also simply by "seeing" that they are isomorphic seems a little bit like a "hand-waving" argument. So maybe someone has an elegant and more sound solution? I have no other ideas?


You can try to solve it using generators and relations. Your group is basically $$<x,y \mid \; x^{p^a}=y^{p^b}=1,\; x^{p^{a-n}}=y^{p^{b-n}}, [x,y]=1>.$$ From your choice of $U_1,U_2$ you actually get that $y^{p^b}=(y^{p^{b-n}})^{p^n}=(x^{p^{a-n}})^{p^n}=x^{p^a}$, so we can remove the relation $x^{p^a}=y^{p^b}$ without changing the group.

Assume wlog that $a\geq b$ and we consider only the relations $$<x,y \mid \; x^{p^a}=1,\; x^{p^{a-n}}=y^{p^{b-n}}, [x,y]=1>.$$ The trick now is to change the basis of your groups. $$1=x^{p^{a-n}}y^{-p^{b-n}}=(x^{p^{a-b}})^{p^{b-n}}y^{-p^{b-n}}=(x^{p^{a-b}}y^{-1})^{p^{b-n}}$$ Setting $z=x^{p^{a-b}}y^{-1}$ we get that the group is actually $$<x,z \mid \; x^{p^a}=1,\; z^{p^{b-n}}=1, [x,z]=1>\cong C_{p^a}\times C_{p^{b-n}}.$$

  • $\begingroup$ How you see that $x, z$ generate the group? In the original group every element is of the form $x^n y^m$ with $0 \le n < p^a, 0 \le m < p^b$, but how to write this as a product of $z$ and $x$? $\endgroup$ – StefanH Jan 3 '16 at 16:15
  • $\begingroup$ Okay, I see, simply note that $y = z^{-1} x^{p^{a-b}}$ and so it generates the same group. I like your solution, but generators and relations appear some chapters later in the book were this exercise is taken from. So if anybody has an alternative solution I would be interested in. $\endgroup$ – StefanH Jan 3 '16 at 16:43
  • $\begingroup$ @Stefan I think that generators and relations is the easiest way to present this solution. The main idea here was to choose a new "basis" for the group and use the relations in order to figure out the right basis to choose. $\endgroup$ – Ofir Jan 3 '16 at 16:54

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