I am trying to do an exercise but I am confused at this question. Isn't the order of $a$ just $3$.

I am trying to use Lagrange Theorem but to no avail since I have no clue what $G$ is.

Order Group Question

  • $\begingroup$ Well, at least $\;30\;$ , uh? $\endgroup$ – DonAntonio Dec 12 '17 at 0:31
  • $\begingroup$ @DonAntonia How did you get that ? I understand that 3 divides the order of the factor group and order of G divides 10 thus 30 is number of elements in G at least but how does that pertain to order of a ? $\endgroup$ – Nameless Dec 12 '17 at 0:32
  • $\begingroup$ So, offhand, you know that $a^3H = H$, not that the order of $a$ is $3$. $\endgroup$ – Ted Shifrin Dec 12 '17 at 0:33
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    $\begingroup$ @DonAntonio - what about $a =(1,0)\in {\mathbb Z/3}\times {\mathbb Z/10}$? $\endgroup$ – peter a g Dec 12 '17 at 0:34
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    $\begingroup$ Let $m$ be the order of $a$.Then $(aH)^{m} = a^{m}H = H$. It follows that $3|m$ so that $m = 3k$. $\endgroup$ – akech Dec 12 '17 at 1:19

$|a|$ clearly divides $30$, for we have:

$a^{30} = (a^3)^{10} = h^{10}$ for some $h \in H$, and by Lagrange, $h^{10} = e$ (since $\langle h\rangle$ is a subgroup of $H$ and its order, which is $|h|$, must divide $10$).

On, the other hand, if $|a| = k$, then $a^k = e \in H$, and since $aH$ has order $3$, we must have $3 \mid k$ (for otherwise, $aH = H$, or $a^2H = H$, contradiction).

The possibilities are thus:


Letting $G = \Bbb Z_{30}$ and $H = \langle 3\rangle$ (which has order $10$), we see that $1+H$ and $2+H$ both have order $3$ in $G/H$, and:

$10 \in 1+H$ has order $3$ in $G$.

$5 \in 2+H$ has order $6$ in $G$.

$2 \in 2+H$ has order $15$ in $G$, and

$1 \in 1+H$ has order $30$ in $G$, which settles the question.


Consider a subgroup $K\subseteq G/H$ generated by $aH$. It follows that $K\simeq G'/H$ for some subgroup $H\subseteq G'\subseteq G$ such that $a\in G'$. Since $|G'/H|=3$ and $|H|=10$ then it follows (by Lagrange's theorem) that $|G'|=30$. And since $a\in G'$ then $|a|$ divides $30$.

Are all divisors of $30$ possible? If you consider $G'=\mathbb{Z}_3\times\mathbb{Z}_{10}$ and $H=0\times\mathbb{Z}_{10}$ then $3, 6, 15, 30$ are clearly possible values of $|a|$. Obviously $1$ is not possible.

What about $2, 5$? (thanks to @DavidWheeler) Of course if $|a|=2$ or $|a|=5$ then $K$ being the image of $\langle a\rangle$ would have to be of order $1,2$ or $5$. But $|K|=3$. Contradiction.

  • $\begingroup$ 2 is not possible, for if so, $aH$ could not have order 3. 5 is likewise not possible for it leads to $H = a^3H = a^5H \implies a^2 \in H$. $\endgroup$ – David Wheeler Dec 12 '17 at 12:06

A variation on David W's answer.

Write $\langle a\rangle$ for the cyclic subgroup of $G$ generated by $a$.

One has a surjective homomorphism $\pi$

$$ \left< a \right> \to \langle a\rangle H / H,$$ defined by $\pi\colon a^k\mapsto a^k H$. This is a homomorphism, as $H$ is normal.

The kernel $K$ of $\pi$ is $\langle a\rangle \cap H$. By hypothesis the image of $\pi$ has cardinality $3$. Therefore $\#\langle a \rangle = 3 \ \cdot \#K$. Now, the only possibilities for the cardinality of $K$ are $1,2,5, 10$, as $K$ is a subgroup of $H$. So the only possibilities for the order of $a$ are $3, 6, 15, 30$. By explicit construction (as in the comments, or in David's answer), we see all of these occur.


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