If $G, H$ are groups and $G$ is isomorphic to $H$, prove the following:  If $G$ has an element of order $n$, then $H$ has an element of order $n$, and  If each element of $G$ has finite order, then each element of $H$ has finite order.    For both, isn't there a theorem that states the exact thing? If this is straight definition, what am I suppose to prove?
9546
Let's say $f$ is an isomorphism $f:G  \rightarrow G'$, where $G$ and $G'$ are multiplicative groups.  Then for $x\in G$, if $f(x) = x' \in G'$. Do we have always have $\text{ord}(x) = \text{ord}(x')$? Why?
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Show that any cyclic group of even order has exactly one element of order $2$   Attempt:  Let $G$ be a finite group of even order.  Assuming a set $A=\lbrace g \in G \vert g \neq g^{-1} \rbrace$, I can show that  $A\cup\{e\}$ has odd number of elements as $A$ has even number of elements. So $G$ has at least one element $a$ such that $a$ does not in $A$, ie $a=a^{-1}$ ie $a^2=e$. That is $G$ has atleast one element of order $2$. "More specifically, $G$ has odd number of elements of order $2$ --- ???"  But how to show that exactly one such element if $G$ be cyclic.     NB: Please don't use isomorphic properties.
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Show that cyclic group of finite order $n$ is isomorphic to $(\mathbb Z_n,+)$ Actually, the answer to this question is available in MSE but I am not getting it,and one answer is not in the English language.Please kindly give a proof in easy language. Thank you.\ Here is a solution I don't understand:
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I would like to prove that : If $H$ is a subgroup of $G$ and $K$ is a subgroup of $H$, then  $$|G:K|=|G:H||H:K|$$  I appreciate any clear explanation of this. Thanks !
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Let $f$: $Z_m \rightarrow G$, $f(n)$ = $a^n$. Show that f is a function and an homomorphism.  Doing a review on the whole semester, and a little refresh on this topic would be great.   By definition I know $G$ has the same order of $Z_m$, but not sure how to attack this problem with ease.   I also know that cyclic groups of the same order are isomorphic.   (I cannot use the fact that cyclic groups of the same order are isomorphic, that's why I ask for another approach, because I can't use theorems or lemmas that have not been discussed in class.)
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Let $H$ and $K$ be subgroups of $G$. Let $H$ have order $m$ and $K$ have order $n$, where $m$ and $n$ are   relatively prime. Then  $H \cap K=\{e\}$   My proof:  Let $H$ and $K$ be subgroups where the $ord(H)=m$ and $ord(K)=n$ and $m,n$ are relatively prime. We know that $H \cap K$ is a subgroup of $H$ and $K$ since it contains the elements of both in $H$ and $K$. We shall let the ord($H \cap K$)=d.   Then, by Lagrange's theorem, the order of $H$ is a multiple of the order of $H \cap K$. In other words, $m$ is a multiple of $d$ or $d|m$. Similarly,  by Lagrange's theorem, the order of $K$ is a multiple of the order of $H \cap K$. In other words, $n$ is a multiple of $d$ or $d|n$. Since $d$ divides both $m$ and $n$ and we know $m$ and $n$ are relatively prime then, the order of $H \cap K$ must be $1$.   Since $H \cap K$ is a subgroup, then by properties of subgroup, it contains an inverse which means it must also contain an identity and since $H \cap K$ contains one element, it must contain the identity. As a result, $H \cap K=\{e\}$  Hopefully, someone can confirm or correct any mistakes I made. Thanks!
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Theorem: If $G$ is a finitely generated group, then it has a finite (maybe zero) number of subgroups of index $n$ for any $n\in \mathbb{N}$.   is a sketch of a proof. It consists of assuming such a subgroup $H$ exists, and allowing $G$ to act on $G/H$ by left multiplication. Such action defines a homomorphism $$\phi :G \rightarrow \text{Sym}(G/H)\cong \mathbb{S}_n$$  If $G=\langle g_1, \ldots ,g_r\rangle$, then the homomorphism is uniquely determined by where it sends the generators $g_i$. In particular, there are at most $(n!)^r$ such homomorphisms.  The last step of the proof requires one to notice that a different subgroup $H'$ of index $n$ in $G$ would produce a different homomorphism, and here is where I'm struggling. I've been able to show that $\text{ker}(\phi )=\text{core}(H)$, yet haven't been able to rule out the possibility of $\text{core}(H)=\text{core}(H')$.
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I need help to prove this result:  "Let $G$ be a group such that the intersection of all its subgroups other than $\{1\}$ is a subgroup different from $\{1\}$. Then all its elements have a finite order".  I know I must think of an element $g$ of infinite order. That will imply that every subgroup has infinite order (because it will contain an element like $g^k$). After this, I don't know which step I can take. Can someone help?
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Let $A$ and $B$ be finite abelian groups. Suppose that for every natural number $m$, the number of elements of order $m$ in $A$ is equal to the number of elements of order $m$ in $B$. Prove that $A$ and $B$ are isomorphic.  Idea Given that these groups are finite, I think you have to use the primary decomposition theorem somehow.
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Let $G$ be a finite group and suppose there exists $f\in\text{Aut}(G)$ such that $f^2=\text{id}_G$, i.e., $f$ is its own inverse, and such that $f$ has no fixed points other than the identity $e$ of $G$, i.e., $f(x)=x\Rightarrow x=e$. Show that $G$ is necessarily abelian.   While trying to do this exercise I noticed two facts.  First, $g$ and $f(g)$ have the same order because $o(f(g))|o(g)$ and, applying $f$ again and using $f^2=\text{id}_G$, $o(f(f(g)))=o(g)|o(f(g))$ and, once the order of any element is $\geq 1$, it follows that $o(g)=o(f(g))$. Also, it's easy to see that $g$ and $f(g)$ commute.  Second, there cannot exist such an automorphism if the order of $G$ is even, because $f(e)=e$ and we can form pairs like $\{g,f(g)\}$ with $f(g)\neq g$ that are invariant under $f$, i.e., $f(\{g,f(g)\})=\{g,f(g)\}$. But once the order of $G$ is even, proceeding with the construction of the pairs, we'll end up with just one element $\neq e$, so we must have some $\gamma\in G\setminus{e}$ with $f(\gamma)=\gamma$, contradicting the hypothesis.  Let $n$ be the order of $G$ and let's fix an enumeration of the elements of $G$, say $G=\{g_1,\ldots,g_n\}$. My approach was the following. For each $\sigma\in S_n$, let $x_{\sigma}=\prod_{i=1}^ng_{\sigma(i)}$. Then we have that $f(x_{\sigma})=x_{\sigma^{\prime}}$. If it's shown that $f(x_{\sigma})=e\;\forall\;\sigma\in S_n$, then we'd have $x_{\sigma}=e\;\forall\;\sigma\in S_n$, which implies that $G$ is abelian. The problem is that $S_n$ has $n!$ elements while $G$ has $n$ elements, so there repetitions among the $x_{\sigma}$, and we cannot apply directly the reasoning of the last paragraph.  Is this the right way, or is there an easier manner to solve this?
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I was trying to prove this question.   Q. If G is a group of even order, prove that it has an element a, such that a$\neq$e satisfying $a^2$=e.   I tried solving it using Principle of mathematical induction.  So, Let O(G)=2m, where m$\in$ $N$.   We proceed by induction on m.   For m=1, O(G)=2, $\therefore$ G contains two elements.  G=$\{{e,a}\}$. Now, e has order 1. So, a can not have order 1. $\therefore$ a has order 2. So, it true for m=1.  Now, let it be true for m=k.  Then, we have a group G of order 2k, where $\exists$ an element a$\in$ G, whose order is 2.   We wish to prove it for m=k+1.   Then G has order 2k+2.   I don't know how to proceed further.
123633
I was thinking about the following problem:  Suppose that $G_1 \cong G_2$ are isomorphic groups. Under what conditions on the groups $H_1,H_2$ will we have $$G_1 \times H_1 \cong G_2 \times H_2 ?$$  Obviously, in the finite case we must have $|H_1|=|H_2|$. Also, it is easy to see that $H_1 \cong H_2$ is sufficient. Is it necessary as well?  Thank you!
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Let $G$ be a finite abelian group of order $n$ and let $H$ be a subgroup of $G$ of order $m$. Show that $G$ has a subgroup isomorphic to $G/H$.   Here are my thoughts:  Define $\mu_n := \{z \in \mathbb{C}|z^n=1\}$ and $\mu:=\bigcup_{n=1}^{\infty}\mu_n$.  Since a finite abelian group is self-dual, we have $G/H \cong \operatorname{Hom}(G/H,\mu)$.   By the structure theorem for finite abelian groups, we have $$G \cong \mathbb{Z}_{n_1}\times \cdot \cdot \cdot \times \mathbb{Z}_{n_s}$$ where $n_{i+1}|n_i$ for $1 \leq i \leq s-1$ and $n_1\cdot \cdot \cdot n_s=n$, and $n_i \geq 2$ for all $i$.  Also, we have $$H \cong \mathbb{Z}_{m_1}\times \cdot \cdot \cdot \times \mathbb{Z}_{m_l}$$ where $m_{i+1}|m_i$ for $1 \leq i\leq l-1$ and $m_1 \cdot \cdot \cdot m_l=m$, and $m_i \geq 2$ for all $i$.   I want to somehow write $G/H$ as a product of cyclic subgroups of the $\mathbb{Z}_{n_i}$, for now denote this product by $A_1\times \cdot \cdot \cdot \times A_s$, where $A_i \leq \mathbb{Z}_{n_i}$, so that then I can write $$\operatorname{Hom}(G/H,\mu) \cong \operatorname{Hom}(A_1, \mu)\times \cdot \cdot \cdot \times \operatorname{Hom}(A_s, \mu).$$  Then, since $\operatorname{Hom}(A_i,\mu) \leq \operatorname{Hom}(\mathbb{Z}_{n_i},\mu)$ and (by an already known result) $\operatorname{Hom}(\mathbb{Z}_{n_i},\mu) \cong \mathbb{Z}_{n_i}$, I will have $$\operatorname{Hom}(G/H, \mu) \leq \mathbb{Z}_{n_1}\times \cdot \cdot \cdot \times \mathbb{Z}_{n_s}. $$  My problem is I cannot see how to write $G/H$ as such a product or if it's even possible. I appreciate any guidance and suggestions. Thanks.
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Let $G$ be a group and $H_1$ and $H_2$ be two subgroups of $G$. Is it true that if $H_1$ is isomorphic to $H_2$ then $G/H_1$ is isomorphic to $G/H_2$? If that is not true then in what conditions that holds? Any help will be useful.. Thanks
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Prove that any two cyclic groups of the same order are isomorphic.   Let the groups be $G,H$ with order $k$. Let $G=&lt;a&gt;$ and $H=&lt;b&gt;$. Thus we have $|a|=|b|=k$ and by definition, $G=\{a^0,a^1,...,a^{k-1}\}$ and $H=\{b^0,b^1,...,b^{k-1}\}$. Next I set up the bijection $\theta : G \rightarrow H: \theta(a^n)=b^n$. I think I should proceed by showing $\theta$ is an isomorphism but I'm not sure how to do that. What's a good way to go about showing $\theta$ is an isomorphism?
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I want to prove at any group of order $4$ is isomorphic to either $\Bbb Z_4$ or $\Bbb Z^*_8$. I know that these two groups are not isomorphic to each other because they have different order, but I cannot go further to prove at any group of order 4 should be isomorphic to either of them. Help would be greatly appreciated.  Thanks!
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Problem  Let $n \geq 3$ be an odd number and let $G=\{1,...,2n\}$ be a group of order $2n$. Let $\phi:G \to S_{2n}$ be the morphism defined by $\phi(g_i)(g_j)=g_ig_j$ and let $H=\phi^{-1}(A_{2n})$. Show that $H$ is a proper normal subgroup of $G$.  So basically the result of this problem is that every group of order $2n$, with $n$ odd, is not simple.   The morphism $\phi$ can be seen as the action of $G$ on itself by left multiplication. Since $A_{2n} \lhd S_{2n}$, then $H=\phi^{-1}(A_{2n}) \lhd G$. I am stuck at showing that $H \neq \{1\},G$. All I could think of is that $G$ has an element $g$ of order $2$, then $\phi(g)$ is of order $1$ or $2$, but it is easy to see that $\phi(g)=1$ if and only if $g=1$, so $ord(\phi(g))=2$. I don't know which are the elements of order two in $S_{2n}$. I would appreciate some help to solve the problem, thanks in advance.