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Show that a group with $|G| = 33$ contains an element of order $3$.

Using Lagrange's theorem, the possible orders are $1$,$3$,$11$, and $33$ (from the factors of $33$).

If there is a subgroup containing an element with an order of $33$, then we are done.

There is only one identity element with an order of $1$.

I am currently left with elements of orders $3$ or $11$. What is the best way to show that any element of order $11$ is a contradiction?

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  • $\begingroup$ do you know already about the Sylow's Theorems? $\endgroup$ – janmarqz Feb 3 '14 at 4:33
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    $\begingroup$ @janmarqz Cauchy's theorem alone should suffice. $\endgroup$ – Henry Swanson Feb 3 '14 at 4:34
  • $\begingroup$ Why does the fact that an element of order 33 concludes the proof? Is there a typo? $g^{33}=g^{3}$? $\endgroup$ – Mathematicing Mar 25 '17 at 8:47
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If $x$ has order $1$, then $x^1 = e$. But that implies $x = e$, so it must be the identity.

You don't need to show that no element has order $11$, but that all elements having order $11$ is a contradiction.

Assume all non-identity elements of $G$ have order $11$. Let $x \in G$, and consider $\langle x \rangle$. It has order $11$, so there's some $y \in G$ that's not in it. Because the intersection of two subgroups is a subgroup, and $11$ is prime, we use Lagrange's theorem to show $\langle x \rangle \cap \langle y \rangle = \{e\}$.

We claim that for $0 \le i,j < 11$, $x^i y^j$ are distinct. If $x^a y^b = x^c y^d$, then $x^{-c} x^a y^b y^{-d} = x^{-c} x^c y^d y^{-d}$, which reduces to $x^{a - c} y^{b - d} = e$. This implies that $y^{b - d} \in \langle x \rangle$, but since the intersection was trivial, $y^{b - d} = e$, and so must $x^{a - c}$. But that means $a \equiv c \pmod{11}$ and $b \equiv d \pmod{11}$, and so the $x^i y^j$ were distinct.

But that's way too many elements! That's $121$ elements, but there were only $33$ in $G$. By contradiction, some element must have order $\ne 11$, and you've already shown the only other choice is $3$.

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Suppose that all the non-identity elements in $G$ have order $11$. Choose an $x_1 \in G, x_1 \ne e$; then $H_1 = \langle x_1 \rangle$ is a subgroup of order $11$.

Now choose $x_2 \in G - \langle x_1 \rangle$; $x_2$ has order $11$, and we again form $H_2 = \langle x_2 \rangle$. Now as a consequence of Lagrange's theorem, $$H_1 \cap H_2 = \{e\}$$

(there is only one proper subgroup of either $H_1, H_2$); thus $H_1 \cup H_2$ account for $11 + 11 - 1 = 21$ elements in the group.

Rinse and repeat to get $H_3$. Argue as before to conclude that this is almost disjoint from both $H_1$ and $H_2$. This accounts for $11 - 1 = 10$ more elements, and a total of $31$.

Question: How do we get the remaining two elements?

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According to Lagrange's Theorem, the possible orders for proper subgroups of $\;G\;$ of the form $\;\langle x\rangle\le G\;,\;\;x\in G\;$ are $\;1,3,11\;$ . This number, of course, is also the order of the element $\;x\;$ .

If all the elements of $\;G\;$ had order $\;11\;$ , and since in a cyclic group of order $\;11\;$ there are exactly $\;10\;$ elements of order $\;11\;$ (why?), we'd get at most $\;10+10+10=30\;$ elements (we can't have more elements of order$\;11\;$ since there's one unique element of order $\;1\;$ ...!) .

Thus, there must be at least one element of order $\;3\;$.

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  • $\begingroup$ I don't see where $11+11$ comes from, but that's OK, it gives OP something to correct. $\endgroup$ – Gerry Myerson Oct 15 '13 at 11:56
  • $\begingroup$ From two different cyclic subgroups of order $\;11\;$. Continuing the argument shows that there can't be three of them. In fact it should be $\;10+10+10=30\;$ ... $\endgroup$ – DonAntonio Oct 15 '13 at 11:57
  • $\begingroup$ Too bad. Now there's nothing left for OP to do (except copy-paste your answer into his homework). $\endgroup$ – Gerry Myerson Oct 15 '13 at 21:53
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Suppose there is no element of order $3$; then every element has order $1$ or $11$. Define an equivalence relation on $G$, calling two elements $a$ and $b$ equivalent if $\langle a\rangle=\langle b\rangle$. The equivalence class of the identity element contains only that one element. The equivalence class of an element of order $11$ contains exactly $10$ elements. Thus the non-identity elements of $G$ are partitioned into disjoint $10$-element sets. Since $32$ is not divisible by $10$, this is impossible.

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Teacher's solution: Let $G$ be a group of order $33$. By Lagrange's Theorem, $o(x) \in \{1,3,11,33\}$.

If $o(x)=1$, then the element must be the identity $e \not= x \in G$. We can rule out the order of $1$.

If $o(x)=3$, then we are done. So suppose $o(x) \not=3$. Then either $o(x) =11$ or $o(x)=33$.If $o(x)=33$, then $o(x^{11})$ is an element of order $3$ (as $o(x^{11})^3 = o(x^{33})= e$). So suppose that $o(x) \not= 33$ as well. Then $o(x) = 11$.

Assume that all non-identity elements $x \in G$ have order of $11$. So we have $\langle x : x^{11} = e \rangle \le G$. Let $\langle y \rangle = G \setminus \langle x \rangle$. If $\langle x \rangle \cap \langle y \rangle \le \langle x \rangle$, then $\langle x \rangle \cap \langle y \rangle = \{ e \}$. By Lagrange's theorem, $| \langle x \rangle \cap \langle y \rangle | \mid |\langle x \rangle|=11$. If we list all the elements of $\langle x \rangle$ and $\langle y \rangle$, we would only get $21$ elements, which is not enough. But if we were to add in generators $\langle z \rangle$ and $\langle w \rangle$, then we would get $41$ elements, which is too many. So all elements of order $11$ will yield a contradiction.

Therefore, $G$ must contain an element of order $3$.

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Hint:

Since $\lvert G\rvert\neq 1$, there is certainly some element $x\in G$, $x\neq 1$. Consider $H$, the group generated by $x$.

By Lagrange's Theorem (do you have this result yet?) you know that $\lvert H\rvert$ must divide $\lvert G\rvert=33$. So, there are four cases: $\lvert H\rvert$ must be $1$, $3$, $11$, or $33$.

It can't be the case that $\lvert H\rvert=1$. What happens in the other cases?

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This could very well be treated in general. Whenever there is a prime $p$ that divides the order of $G$, then the number of elements of order $p$ in $G$ is a multiple of $p-1$.

$11$ is prime. If all nonidentity elements of $G$ are of order $11$, then the number of these elements in $G$ is a multiple of $10$. We got a contradiction because $|G|=10\cdot3+2+1$.

The best hint here is that of @daniel-fisher .

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First of all, if there's a non-identity element that is of order one, then you have 2 distinct identities, which must be identical. Contradiction.

Now to the main point, look at an element $g \in G$ and the cycle $\langle g \rangle$ it generates. That must have order $|\langle g \rangle| = 3, 11,$ or $33$ since we can safely avoid the cycle of order $1$. If we have $3$ then we're done. If it has order $33$ then $G = \langle g \rangle$ which means $g^{11}$ has order $3$. But if $|\langle g \rangle| = 11$ then by Lagrange's Theorem $|G/ \langle g \rangle| = 3$. This is vacuously cyclic so choose one of the generating elements. It has order $3$.

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    $\begingroup$ We cannot be certain $\langle g \rangle$ is normal in $G$ without more advanced machinery. $\endgroup$ – Henry Swanson Feb 3 '14 at 4:51

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