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I have the direct product: $G_{1} \times G_{2}$ and I know it is a cyclic group. I want to prove $G_{1}$ and $G_{2}$ are themselves cyclic groups. Any help getting started?

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  • $\begingroup$ Try some simple examples and discover what's going on. $\endgroup$ – Ted Shifrin Oct 22 '17 at 2:40
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    $\begingroup$ Let $G$ and $H$ be groups, and $f:G\to H$ surjective homomorphism. If $G$ is a cyclic group, then $H$ is cyclic also. $\endgroup$ – M. Wolf Oct 22 '17 at 3:07
  • $\begingroup$ In your case, you can consider the homomorphisms $\pi_i:G_1\times G_2 \to G_i$, where $i=1,2$. $\endgroup$ – M. Wolf Oct 22 '17 at 3:09
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    $\begingroup$ Subgroups of cyclic groups are cyclic. $\endgroup$ – Randall Oct 22 '17 at 3:17
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Your group $G_1 \times G_2$ is cyclic, i.e. there exists $g \in G_1 \times G_2$ which generates the big group. Be aware that $g = (g_1, g_2)$ for some $g_1 \in G_1$ and $g_2 \in G_2$. So why not trying to prove your statemant with these two "obvious" candidates?

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  • $\begingroup$ Could we then say that there exists subgroups of $G_{1} \times G_{2}$, H = $G_{1} \times \{e_{2}\}$ and J = $\{e_{1}\} \times G_{2}$. And since we know that the subgroups of cyclic groups are themselves cyclic, H and J are cyclic?? $\endgroup$ – a.powell Oct 22 '17 at 16:01
  • $\begingroup$ yes, that's correct. $\endgroup$ – M.U. Oct 26 '17 at 7:29
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Any subgroup of a cyclic group is itself cyclic (do you know why?). Then you can identify $G_1$ with $G_1 \times \{1\}$ and the same with $G_2$.

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Here are some thoughts. Working with the integers $\mod p$ where $p$ is a prime number, if $G$ is a cyclic group of order $p-1$, and $x$ is a primitive root ($x$ has multiplicative order $p-1$), and $G=[x,x^2,x^3,...x^{p-1}]$, then every non-zero element $e$ is contained in $G$. This is the "bigger group". Let $a$ and $b$ be relatively prime integers such that $ab=p-1$. If $y$ is an element with multiplicative order $a$, and $z$ is an element with multiplicative order $b$, the cyclic group of order $a$ is $A=[y,y^2,y^3,...y^{p-1}]$ and the cyclic group of order $b$ is $B=[z,z^2,z^3,...z^{p-1}]$ are the "smaller groups" and hence subgroups of $G$. It follows the multiplication of every element in $A$ with every element in $B$ will yield the group $G$. That is Group $A$ multiplied by group $B$ is group $G$, or $AB = G$. Another consequence is that every non-zero element $e$ is the product of an element in $A$ and $B$. In other words, each non-zero element $e$ has the form $y^nz^m$ for integers $n$ and $m$ where $y$ and $z$ are defined earlier.

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  • $\begingroup$ I don't see how this in any way answers the question asked. $\endgroup$ – Tobias Kildetoft Jan 19 '18 at 7:16

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