Show that a homomorphism of groups also has the property that $f(a^{-1})=f(a)^{-1}$ for all $a \in G$.
121422
Let $G_1$ and $G_2$ be groups and $\phi:G_1\to G_2$ be a group homomorphism. Show $\phi(g^{-1})=(\phi(g))^{-1}$ $\forall g_1\in G_1$.  So far I have:  $\phi(e)=e$ by definition, where $e$ is the identity element. Then $\phi(g_1)\phi(g_1^{-1})=\phi(g_1g_1^{-1})=\phi(e)=e$ since $\phi$ is a group homomorphism.  I know I can get $\phi(g^{-1})=(\phi(g))^{-1}$ $\forall g_1\in G_1$ from that, but I don't know HOW.  Any help/hints would be welcomed. ^_^
122411
Let $G = (G,\times)$ be a group, and $G^{op} = (G,o), a \ o \ b = b \times a$ be it's opposite group.  I need to prove that there is a group isomorphism $f: G \to G^{op}$ Obviously, $id_G: G \to G$ is a bijection, but it's a group homomorphism if and only if $G$ is commutative.
60423
If $H$ be a subgroup of a group $G$  such that $Ha \not=Hb$ implies that $aH\not=bH$.Then how can I show that $gHg^{-1}\subset H$  $\forall$ $g\in G$?  I do not think I have made any progress.However, this is what I have done: For any $h\in H$,$(ghg^{-1})(gh^{-1}g^{-1})=e$ where $e$ is the identity element of the group $G$.  And for $h,k\in H$, $(ghg^{-1})(gkg^{-1})=g(hk)g^{-1}$ which is in $gHg^{-1}$ as $hk\in H$. Thus $gHg^{-1}$ is a group.  I do not think I am getting anywhere.
113212
True or false. (Prove or give a counterexample.) Let $G$ be a group. Then $|g| = |\phi(g)|$, for all homomorphisms $\phi: G \to G$ and all $g \in G$. Solution. False. $\phi: \mathbb Z_{10} \to \mathbb Z_{12}$ defined by $\phi(x)=0$ for all $x \in \mathbb Z_{10}$ is a counterexample. This function is a homomorphism because $\phi(x+y) = 0 = 0+0 = \phi(x) + \phi(y)$ for all $x, y \in \mathbb Z$ (this function is discussed in problem 3 in assignment 7 as the function sending $[1]$ to $[0]$). The order of $g=1$ is infinity. The order of $\phi(1)=0$ is one.  The problem is that he said $G \to G$ is equivalent to $\mathbb Z_{10} \to \mathbb Z_{12}$, which is think is not right. Explain please? EDIT: Please look at this test and give me your honest opinion,
5791
I'm looking for verification:  My claim: If $G$ is a finite group and $H$ is a (proper)subgroup of index $k&gt;1$, where $k! \leq |G|$, then $G$ is not simple.  Proof: Consider the set of left cosets of $G$ by $H$:  $$G/H = \{H,g_1H,\dots,g_kH\}$$  Then $\phi: G \rightarrow S_{G/H}$ is a homomorphism where we define:  $$\phi(g)[g_iH] = g g_iH$$  ($S_{G/H}$ is the group of permutations on elements of $G/H$ and of course $S_{G/H} \cong S_k$.)  Also $\ker \phi \neq G$ because $\phi(g_1)$ is clearly not the identity in $S_{G/H}$.  Because $\phi$ is a homomorphism,  $$ G/\ker\phi \cong \text{im}\phi$$  Then $|\text{im} \phi| \leq |S_k| = k!$ and if $k! \leq |G|$ then $|\ker\phi| &gt; 1$ and $\ker\phi$ is a non-trivial normal subgroup of $G$.  Example: If a group of order $36$ has a subgroup of index $3$ or $4$ then the group is not simple.
77790
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 $\text{core}(H)=\text{core}(H')$.
38710
Let $G$ be a group and $H$ a subgroup of $G$. For $x \in G$, prove $x^{-1}Hx$ is a subgroup of $G$.
19796
Let $\phi: G \to H$ be a group homomorphism, and let $g$ be an element of $G$. Show that the order of $\phi(g)$ divides the order of $g$.
18867
I am working on a school assignment and have been stuck on this question for some time.   Let $f : G \rightarrow G_1$ be a surjective homomorphism (also called epimorphism) from $G$ to another group $G_1$. Prove that $f(Z(G)) \subseteq Z(G_1)$  I know we start by assuming we have an element in $f(Z(G))$ then showing it is also in $Z(G_1)$. Also, $Z(G)$ is the center of $G$, so $Z(G)$ is all the elements of $G$ that commute with all elements in $G$. But how do we know that if we map the elements of $Z(G)$ to $G_1$, then their image will also commute with all the elements in $G_1$? Does it have something to do with using commutativity with the homomorphism? i.e. $f(gz) = f(g)f(z) = f(zg) = f(z)f(g)$.
72914
If $f: G \to H$ is an isomorphism of groups then show that for any integer $k$ and $g \in G$, these sets $A = \{a \in G : a^k = g$} and $B = \{b \in H : b^k = f(g)\}$ have the same number of elements.
96801
Let $G$ and $K$ be groups and $f:G\rightarrow K$ be a group   homomorphism. Suppose $g\in G$ has finite order $n$. Prove that   $\mathrm{ord}(f(g))$ is finite and divides $n$.   How can I fix this proof?  We know $g^n=e$. Since $f$ is a homomorphism, then $[f(g)]^n = f(g^n)=f(e)=e$. Let $k\mid n$ and $j\mid n$. Then we can choose the smallest $k$ such that $[f(g)^k]^j=[f(g)]^n$. Thus, $\mathrm{ord}(f(g))=k$ where $k\mid n$.  This logic seems shaky. How can I improve it? Thanks!
23648
I am trying to prove a series of propositions:  Given any homomorphism p from $\mathbb{R}$[X] to $\mathbb{R}$[X], show that it is equal to $\phi_g$ for a unique g in $\mathbb{R}$[X], with $\phi_g$(f) = f(g(X)). I've expanded the expression $p(f) = p(\sum_{i=0}^n a_iX^i) =\sum_{i=0}^n p(a_i)p(X)^i$ but I'm not sure how to show $p(f) = \sum_{i=0}^n p(a_i)p(X)^i = \sum_{i=0}^n a_ig(X)^i = \phi_g(f)$.  Show that if h,g $\in \mathbb{R}$[X] are such that h(g(X)) = X, then g(X) = aX+b for a $\in$ R$^x$ and b$\in \mathbb{R}$.
118745
Let $f:G\to K$ be a surjective group homomorphism and $H$ be a subgroup of $G$ satisfying $N_G(H)=H$. Then $N_K(f(H))=f(H)$. I'm trying to prove or disprove it. perhaps, this proposition is true. But, how can we prove it? Edit: I thought it is true because WIKI said it is true without proof
85131
If $f: \mathbb Q\to \mathbb Q$ is a homomorphism, prove that $f(x)=0$ for all $x\in\mathbb Q$ or $f(x)=x$ for all $x$ in $\mathbb Q$.  I'm wondering if you can help me with this one?
44724
Let $G$ be a group, and suppose that $a= {a^{-1}}$ for every $a\in G$. Prove that $G$ is Abelian.    I know that I need to prove that ${a^{-1}}$ = a and use right multiplication. I also know that I need to declare a, b $\in$ G and go from there, but I'm confused when I get to   (${a^{-1}}$${b^{-1}}$) = ab.
63316
Let $\phi:G\to G'$ be a group homomorphism, and let $N'$ be a normal subgroup of $G'$. Show that $\phi^{-1}[N']$ is a normal subset of $G'$.  My attempt: $\phi^{-1}[N']=\{g\in G:\phi(n)\in N'\}$ $$g' n' (g')^{-1}\in N'$$ $$\Rightarrow\phi^{-1}(g' n' (g')^{-1})=\phi^{-1}(g')\phi^{-1}(n')\phi^{-1}(g'^{-1})$$ Thus $\phi^{-1}[N']$ is normal. Is this thinking correct? I feel like I am missing something.
27695
$F$ is an isomorphism from group $G$ onto group $B$. Prove $F^{-1}$ is an isomorphism from $B$ onto $G$.  I do not know which direction to go in.