This is Pinter 10.H.1.

Let $a$ denote an element of a group $G$.

Let $a$ have order $12$. Prove that if $a$ has a cube root, say $a = b^3$ for some $b \in G$, then $b$ has order $36$.

HINT: Show that $b^{36} = e$. Then show that for each factor $k$ of 36, $b^k = e$ is impossible. [Example: If $b^{12} = e$, then $b^{12} = (b^3)^4 = a^4 = e$.] Derive your conclusion from these facts.

OK, let's begin.

We are given:

$$ \operatorname{ord}(a) = 12 \tag{1} $$ $$ a = b^3 \tag{2} $$

Here are the factors of $36$:

$$ 2, 3, 4, 6, 9, 12, 18 $$

As suggested by the hint above, for each $k$ of these, we need to show that $b^k = e$.

Let's start with $k = 2$. Assume:

$$ b^2 = e \tag{3} $$

Let's start with (2):

$$ a = b^3 $$

$$ a = b^2 b $$

Substitute (3):

$$ a = e b $$

$$ a = b \tag{4} $$

OK, back to (3):

$$ b^2 = e $$

Substitute (4):

$$ a^2 = e $$

This contradicts (1).

So we've disproved the case where $k = 2$. We continue in a similar way for the rest of the factors of $36$ to complete the exercise.

Second approach

Alright. Let's take a different approach for $k = 2$.

Let's start with (2):

$$ a = b^3 $$

This time, we'll raise both sides to the power of $12$:

$$ a^{12} = (b^3)^{12} $$

$$ a^{12} = b^{36} $$

$$ a^{12} = (b^2)^{18} $$

Substitute (3):

$$ a^{12} = (e)^{18} $$

$$ a^{12} = e $$

This conclusion is compatible with (1).

It seems like the second approach which does not contradict (1) should not be possible. What am I missing? Is there a step I'm taking which is invalid?

Exercises 10.H.2 and 10.H.3 are similar to this one and I can use a similar approach to the alternative one above. So I think by understanding this one, it'll clear up issues in those exercises as well.

In 10.H.1, we are told that $\operatorname{ord}(b) = 36$. So we know that for each factor $k$ of 36, we need to show that $b^k = e$ is not true. However in 10.H.2, and 10.H.3, we are not given the order of $b$. So we need to show definitively, for each possible $k$, whether $b^k = e$ is possible or not. So the second approach above is problematic because it doesn't help us rule out certain values of $k$. I'm wondering, what specific aspect of this approach should be avoided in performing these exercises? My guess is, the misstep is at saying "raise both sides to the power of 12".

  • $\begingroup$ If $a^2=e$, then $a^{12}=e$ as well. $\endgroup$
    – Tab1e
    Aug 17, 2019 at 7:52
  • $\begingroup$ Why should it not possible to compatible with $(1)$? This only means this method does not work. $\endgroup$
    – xbh
    Aug 17, 2019 at 7:53
  • $\begingroup$ Just because your specific line of reasoning in (2) didn't reach a contradiction, that doesn't mean that there isn't a contradiction hiding behind some other line of reasoning from the same assumptions. $\endgroup$
    – Arthur
    Aug 17, 2019 at 7:55
  • 1
    $\begingroup$ This deserves more upvotes for its detailed context and attempts. $\endgroup$
    – Shaun
    Sep 8, 2019 at 21:07

2 Answers 2


There are two factors of $36$ missing: $1$ and $36$.

If $\operatorname{ord}b=k$, then $b^k=e$ and therefore $(b^k)^3=e$. But $(b^k)^3=(b^3)^k=a^k$. So, $a^k=e$ and, since $\operatorname{ord}a=12$, this implies that $12\mid k$.

Clearly $b^{36}=e$. Do we have $b^{12}=e$? No, because $b^{12}=a^4\neq e$. By the same argument, $b^{24}\neq e$. Therefore, $\operatorname{ord}b=36$.


If $a^2=e$ then $a^{12} = e$ also holds. But $Ord(a) = 12$, so the lowest power that results in $e$ is $12$, so $a^2 \neq e$ and $a^{12} = e$. So, the second thing is not wrong, but it doesn't proof anything as you haven't shown whether the lower powers equal e or not.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .