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Problem 1: Let $G$ be an abelian group of order 35, where $x^{35}=e$ for all $x \in G$. Prove $G$ is cyclic.

Problem 2: Let $a,b$ be elements in a group. If $|a|=12$, $|b|=22$ and $\langle a \rangle \cap \langle b \rangle \neq \{e\}$, prove that $a^6 = b^{11}$.

Problem 3: Let $H = \{\alpha \in S_6 : \alpha(2)=2, \alpha(4)=4 \}$. Prove that $H \preceq S_6$.

EDIT: ⪯ means "subgroup of"

I'm having trouble on how to start these problems, ex. for the first question I tried to find an element of order 35 and contra-positive contradiction but can't seem to find the key. Any help would be appreciated!

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    The first step is to type your question instead of using an external link. – 2017-02-07
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    Odd... $x^{35}=e$ for all elements certainly doesn't give you any information that $|G|=35$ doesn't (you've surely had Lagrange's thm by now...). If you have two generators $g_1,g_2$ with orders $l,m$ where $l,m$ are either $5,7$ or $35,$ then the element $g_1g_2$ has order $35.$ It's just how $5$ and $7$ work. – 2017-02-07
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    @spaceisdarkgreen If the order of an element is not $35$, then it's not a generator. Also, you actually have to prove that an element of order $5$ and an element of order $7$ exists, which requires a few sentences. Or Sylow. Also, the way you wrote it, we could have $l = m = 5$, which is probably not what you intended. – 2017-02-07
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    @Arthur I was going at a quasi-contradiction angle with the generators thing but you're right that it's sloppy and wrong in that you would also need to show that there would necessarily be elements of both orders $5$ and $7$ – 2017-02-07
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    If $a$ has order $5$, then so do $a^2,a^3,a^4$, so the number of elements of order $5$ is a multiple of $4$. Similarly, the number of elements of order $7$ is a multiple of $6$. Can you do it using that information? (Note that $34$ is not divisible by $4$ or by $6$.) – 2017-02-07

2 Answers 2

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1: The number of Sylow $5$-subgroups is of the form $1+5k_1$ and divides $7$ so there is just one such; similarly the number of Sylow $7$-subgroups is of the form $1+7k_2$ and divides $5$ so again there is just one. Hence the unioin of all the Sylow subgroups has $1+(5-1)+(7-1)=11$ elements. This leaves $35-11=24$ elements not of order $1$, $5$, or $7$; by Lagrange they must have order $35$ and therefore generate the whole group, i.e. $G$ is cyclic.

2: If none of $a,a^2,\ldots,a^{11}$ is in $\langle b\rangle$, then $\langle a\rangle\cap\langle b\rangle$ would be $\{e\}$; since this is not the case, there must be an $r$ with $1\le r\le11$ such that $a^r\in\langle b\rangle$. Then $a^{22r}=\left(a^r\right)^{22}=e$ $\implies$ $12\mid22r$ $\implies$ $6\mid11r$ $\implies$ $6\mid r$. It follows that $r=6$. So $a^6=b^s$ for some $s=1,\ldots,21$. Then $b^{2s}=a^{12}=e$ $\implies$ $22\mid2s$ $\implies$ $11\mid s$. Hence $s=11$, i.e. $a^6=b^{11}$.

3: If $\alpha,\beta\in H$ then, given $n\in\{2,4\}$:

(i) $(\alpha\beta)(n) = \alpha(\beta(n)) = \alpha(n) = n$
(ii) $\alpha^{-1}(n) = \alpha^{-1}(\alpha(n)) = (\alpha^{-1}\alpha)(n) = 1_{S_6}(n) = n$

So $\alpha,\beta\in H$ $\implies$ $\alpha\beta,\alpha^{-1}\in H$ i.e. $H$ is a subgroup.

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For 1, if there is an element of order $35$, then it generates the whole group, so it's cyclic. Your job is to show that such an element must exist. You do this by showing that there must be both an element of order $5$ and one of order $7$, and then you take their product.

For 2, examining the possible orders of a certain power of $a$ (using Lagrange's theorem in $\langle b\rangle$ and $a^{12} = e$) will lead you to an answer.

For 3, your book should have a list of three criteria to check that a subset is a subgroup. Follow that list.

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    for 1, I can see that if there are elements of order 5 and 7, then there is an element of order 35, but how do I show that there exist elements of order 5 and 7? – 2017-02-07
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    Only one element can have order $1$, the $34$ that are left have to have orders $5,7$ or $35$. Say they all have order $5$. Then they all generate cyclic subgroups of order $5$. Each such group consists of the identity and four other elements, and two such groups intersect only in $\{e\}$. That means that the $34$ non-identity elements are partitioned into disjoint sets of four by what cyclic group they generate. This is impossible, since there would be two elements left over. So we can't have that all elements have order $5$. An analogous argument works for order $7$. – 2017-02-07
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    **You do this by showing that there must be both an element of order 5 and one of order 7, and then you take their product**, that is not necessarily true, since it is not abelian by hypothesis. Sorry, I misread. – 2017-02-07
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    @ZenoCozeno Yes, $G$ is assumed to be abelian. It's right there in the first sentence of the question. – 2017-02-07
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    @delson1337: That $x^{35} = 1 \quad \forall x \in G$ is irrelevant since for every finite group $G$ we have $x^{\lvert G\rvert} = 1$. – 2017-02-07
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    @delson1337: II you know about [the funcdamental theorem of finite abelian groups](http://diamond.boisestate.edu/~liljanab/MATH508/Groups.pdf) then you can state that $G \cong C_5 \times C_7$. Chosing generators $a$ and $b$ in each group permits you to generate $C_{5\cdot 7}$. If you don't know about this theorem I strongly suggest to study it as it will permit you to solve more complex problems with much less effort. – 2017-02-07