# Math Help - some problems about groups

1. ## some problems about groups

1. prove no two of C_8, C_4 X C_2, C_2 X C_2 X C_2 are isomorphic.C_n is cyclic group of order n.

Give, with justification, a group of order 8 that is not isomorphic to any of those groups.

2. Prove that if m and n are coprime then C_m X C_n is cyclic.
if m and n are not coprime. can it be cyclic?

3. H is a normal subgroup of a finite group G. Which of the followings are true?

(i) if G is cyclic then H and G/H are cyclic.
(ii) If H and G/H are cyclic then G is cyclic.
(iii) If G is abelian then H and G/H are abelian
(iv) If H and G/H are abelian then G is abelian.

4. find all homomorphisms between C_11 to C_14

5. H is a subgroup of G and H not equal to G. Show that there is an element of G which does not belong to any subgroup of the form gHg^(-1) for g belongs to G

2. Originally Posted by szpengchao
1. prove no two of C_8, C_4 X C_2, C_2 X C_2 X C_2 are isomorphic.C_n is cyclic group of order n.
$C_8$ is cyclic while $C_4 \times C_2$ is not and $C_2\times C_2 \times C_2$ is not. Now $C_4\times C_2$ has an element of $4$ while $C_2\times C_2\times C_2$ does not. So all three are non-isomorphic.

Give, with justification, a group of order 8 that is not isomorphic to any of those groups.
Use $Q_8$. It is non-abelian.

3. Originally Posted by szpengchao
2. Prove that if m and n are coprime then C_m X C_n is cyclic.
if m and n are not coprime. can it be cyclic?
No, it cannot be cyclic. Let $d=\gcd(m,n)$. We claim if $(x,y)\in (C_m,C_n)$ then $(x,y)^{mn/d} = (1,1)$. Note, $x^{mn/d} = \left( x^m \right)^{n/d} = 1^{n/d} = 1$. Similarly, $y^{mn/d} = \left( y^n \right)^{m/d} = 1^{m/d} = 1$. Now, finally since $n,m$ are not relatively prime it means $d>1$ and so $mn/d, hence, no element has order $mn$.

4. Originally Posted by szpengchao
3. H is a normal subgroup of a finite group G. Which of the followings are true?
(i) if G is cyclic then H and G/H are cyclic.
$H$ is cyclic because every subgroup of a cyclic group is cyclic. Furthermore, $G/H$ is cyclic because there exists $a\in G$ such that $\left< a \right> = G$ this means $aH$ generates $G/H$.

(ii) If H and G/H are cyclic then G is cyclic.
False. Let $G=S_3$ and $H$ its 3-element subgroup.

(iii) If G is abelian then H and G/H are abelian
Yes. Define the natural map $\pi: G\mapsto G/H$ then $G/H$ is a homomorphic image of an abelian group so it must be abelian. Or you can say $(xH)(yH) = (yH)(xH)$ since $xy=yx$.

(iv) If H and G/H are abelian then G is abelian.
False. Use the same example.

5. Originally Posted by szpengchao
5. H is a subgroup of G and H not equal to G. Show that there is an element of G which does not belong to any subgroup of the form gHg^(-1) for g belongs to G
Maybe I am missing something really easy here. But I can only prove it for finite groups , I hope that is what is being ask for. Say $H_1,H_2,...,H_n$ are subgroups such that $H=\bigcup_{k=1}^n H_k$ is a subgroup. Consider $h=h_1h_2...h_n\in H$ where $h_i\in H_i$ because it is closed. So then $h \in H_1$ without lose of generality. So $h_1^{-1}h = h_2...h_n\in H_1$ and so on until $h_n \in H_1$, permuting these products it means $h_1,...,h_n\in H_1$. Thus, it means $H=H_1$. Now assume $G$ is finite. If every $x\in G$ is contained in the conjugate subgroup $gHg^{-1}$ for some $g\in G$ then it means $\bigcup_{g\in G}gHg^{-1}=G$ this union is finite for we assumed $G$ was finite. But by above this union is simply $g_0Hg_0^{-1}$ for some $g_0\in G$. But that means $|H| = |g_0Hg_0^{-1}| = |G|$ and since $H\subseteq G$ means $H=G$. We have shown that if $H\subseteq G$ has this property (for a finite group) then $H=G$ by above argument.
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Now it would be nice if this union can generalize, i.e. if $G$ is any group then $\bigcup_{i\in I}H_i = H$ for $\{ H_i|i\in I\}$ implies $H_k = H$ for some $k\in I$. But sadly this does not work. Consider $G = \mathbb{Z}_2^{\infty}$ to be an $\aleph_0$-tuple $(a,b,c,...)$. Now let $H_1 = \{ (0,0,...),(1,0,...)\}, H_2 = \{ (0,0,...), (0,1,...)\}, ...$. Then they unionize to $G$. But no one groups contains them all.