Could someone please verify my following proof?     For isomorphism $\phi: G\to H$ of the groups $G$ and $H$, show that $|\phi (a)|=|a|$ for all $a\in G$.   Proof: Let $a\in G$ with $|a|=n$. Then $a^{n}=e_{G}$, the identity of $G$. Let $|\phi(a)|=m$. Then $(\phi (a))^{m}=e_{H}$, the identity of $H$, or $\phi (a^{m})=e_{H}$. Since $\phi(e_{G})=e_{H}$, it must be that $a^{m}=e_{G}$. Then it must be that $m=n$. Therefore, $|\phi (a)|=|a|$ for all $a\in G$.
120983
Prove that if $f : G \rightarrow H$ is a group isomorphism, then for any element $x \in G$ one has $o(x) = o(f(x))$, where $o(a)$ denotes the order of the element $a$.  I feel that this proof should be online somewhere but haven't been able to find anything so would be helpful if anyone could point me in the right direction.
59476
Let $G_1, G_2$ and $G_3$ be groups.  Let $\phi: G_1 \to G_2$ and $\sigma: G_2 \to G_3$ be isomorphisms of groups.  Show that $$\sigma\circ\phi: G_1 \to G_3$$ is an isomorphism.  I understand to prove the composition is a homomorphism (operation preserving) and a bijection.  I need help with notation and how to show the operation preserving, onto and one-to-one.  Operation preserving:  $\sigma\circ\phi(ab)=\sigma(\phi(ab))=\sigma(\phi(a)\phi(b))=(\sigma\circ\phi(a))(\sigma\circ\phi(b))$  Am I on the right track?
76373
Suppose $G$ is a group and $H,K \trianglelefteq G$. If $H \cap K =1$ the $hk=kh$ for all $k \in K, h \in H$?  Here is what I have so far:  Take $h \in H$ and $k \in K$ we wish to show $hk=kh$ or equivalently $hkh^{-1}=k$. Now I know $hkh^{-1} \in K$ since $K$ is normal in $G$ but I don't know how to show that $hkh^{-1}=k$ i.e. it is the element we want it to be not some other element of $K$. I haven't used the fact that $H \cap K=1$ yet so I guess it follows as a consequence of that but I'm not seeing how at the moment.
56228
I want to prove that the direct product $G\times H$ of two groups has a subgroup isomorphic to $G$ and a subgroup isomorphic to $H$.  How I thought to prove is that taking a pair $(g,h)$ from $G\times H$ and then to show that for some $g$ in $G$ there is an isomorphism with the pair $(g,h)$ of $G\times H$. and the same for $h$ in $H$ also.  Is this method correct? Or should I try to show that there is a bijection between the groups and there exists an equivalence relation between the groups?  Please explain how should I proceed.
110521
The problem is as follows:     Let $G$ be a group and $H$ a subgroup of $G$ (i.e., $H \le G$); let $x$ be any element of $G$ (i.e., $x \in G$). To prove that $[G:xHx^{-1}] = [G:H]$.   I am able to justify that the $xHx^{-1}$ is indeed a subgroup of $G$ (i.e., $xHx^{-1} \le G$) and that $|xHx^{-1}| = |H|$.   To prove that $[G:xHx^{-1}] = [G:H]$, it is necessary to provide a bijective function $F$ between the set of (different, left) cosets of $H$ (denoted $S = \{ aH : a \in G \}$) and that of $xHx^{-1}$ (denoted $T = \{ a (xHx^{-1}) : a \in G\} $).   I have tried the function $F(aH) = a(xHx^{-1})$, but failed to show it is bijective. Could someone offer a hint of the feasible function?
31727
This is from Fraleigh's First Course in Abstract Algebra (page 82, Theorem 8.16) and I keep having hard time understanding its proof. I understand only until they mention the map $\lambda_x (g) = xg$. Can someone explain this proof step by step? Thank you so much! Here is the proof:  Let $G$ be a group. We show that $G$ is isomorphic to a subgroup of $S_G$. Define a one-to-one function $\phi: G \to S_G$ such that $\phi(xy)=\phi(x) \phi(y)$ for all $x,y \in G$. For $x \in G$, let $\lambda_x: G \to G$ be defined by $\lambda_x (g) = xg$ for all $g \in G$. The equation $\lambda_x (x^{-1} c) = x(x^{-1} c) = c$ for all $c \in G$ shows that $\lambda_x$ maps $G$ onto $G$. If $\lambda_x (a) = \lambda_x (b)$, then $xa=xb$ so $a=b$ by cancellation. Thus $\lambda_x$ is also one to one, and is a permutation of $G$. We now define $\phi: G \to S_G$ by defining $\phi(x) = \lambda_x$ for all $x \in G$. To show that $\phi$ is one to one, suppose that $\phi(x) = \phi(y)$. Then $\lambda_x = \lambda_y$ as functions mapping $G$ into $G$. In particular $\lambda_x (e) = \lambda_y (e)$, so $xe=ye$ and $x=y$. Thus $\phi$ is one to one. It only remains to show that $\phi(xy) = \phi(x) \phi(y)$, that is, that $\phi_{xy} = \lambda_x \lambda_y$. Now for any $g \in G$, we have $\lambda_{xy} (g) = (xy)g$. Permutation multiplication is function composition, so $(\lambda_x \lambda_y)(g) = \lambda_x (\lambda_y (g)) = \lambda_x (yg) = x(yg)$. Thus by associativity, $\lambda_{xy} = \lambda_x \lambda_y$.
96709
Let $G$ be a group and let $f:G\to G$ be an isomorphism. Let $H = \{ a \in G \; | \; f(a)=a^{-1}\}$ Prove that $H$ is a subgroup of $G$ if and only if $G$ is abelian.  My attempt: I already know that $H$ is a subset of $G$ since $a$ is in $G$. And that $H$ is $a$ group if and only if for every $a$ and $b$ in $H$, $ab^-1$ is in $H$. That is, if $f(a)=a^{-1}$ and $f(b)=b^{-1}$ then $f(ab^-1)=(ab^{-1})^{-1}=(b^{-1})a$.  But I don't know how to derive this nor the converse based on the abelian property of $G$ that for every $a$ and $b$ in $G$, $ab=ba$.  Please help some. Thanks!
30816
As the title suggests. Let $G$ be a group, and suppose the function $\phi: G \to G$ with $\phi(g)=g^3$ for $g \in G$ is a homomorphism. Show that if $3 \nmid |G|$, $G$ must be abelian.  By considering $\ker(\phi)$ and Lagrange's Theorem, we have $\phi$ must be an isomorphism (right?), but I'm not really sure where to go after that.  This is a problem from Alperin and Bell, and it is not for homework.
90216
Suppose that $H$ and $K$ are normal subgroups of a group $G$ such that $HK=G$. I need to show that  $G/(H\cap K)\cong (G/H)\times (G/K)$.  So from the second isomorphism theorem we have that:  $HK/H\cong H/(H\cap K)$ which gives that $G/H\cong H/(H\cap K)$  $HK/K\cong K/(H\cap K)$ which gives that $G/K\cong K/(H\cap K)$  So we must now show that:  $H/(H\cap K)\times K/(H\cap K)\cong G/(H\cap K)$  So consider the map $\theta:H/(H\cap K)\times K/(H\cap K)\rightarrow G/(H\cap K)$ such that $\theta(h(H\cap K)\times k(H\cap K))=hk(H\cap K)$  Now this is a homomorphism and is surjective. To check injectiviy we see that if:  $\theta(h_1(H\cap K)\times k_1(H\cap K))=\theta(h_2(H\cap K)\times k_2(H\cap K))$ then we have that:  $(h_1k_1(H\cap K))=(h_2k_2(H\cap K))$ so that $(h_1k_1)(h_2k_2)^{-1}(H\cap K)=(H\cap K)$ which gives that $h_1k_1(H\cap K)=h_2k_2(H\cap K)$ So we are done.  Is the above correct?  Thanks for any help
45460
Let $G$ be a group and let $a \in G$ be a fixed element. Let $$H= \{\text{$g \in G$ | $g^{-1}ag=a$}\}.$$  Prove that $H$ is a subgroup of $G$.    So I know I need to show:  (i) $H$ is closed under the operation on $G$.  (ii) $H$ is closed under inversion.  (iii) $H\neq \emptyset$    For closure:  Suppose $x,y \in H.$ Then $x^{-1}ax=a$ and $y^{-1}ay=a.$ So  $$\begin{align} (xy)^{-1}a(xy)&amp; = y^{-1}x^{-1}axy \\  &amp; = y^{-1}(x^{-1}ax)y \\  &amp; = y^{-1}ay \\   &amp; = a \\ \end{align}$$ Thus $xy \in H$.    Now how do I show it's closed under inversion? I'm not sure what the inverse is...   And to show $H \neq \emptyset$ all I need to do is show that the identity is in the set. And I'm not sure how to do that either.  Help!
63503
If $\phi$ is a homomorphism from the group $G$ with identity $e$ to the group $G'$ with identity $e'$, then $\phi$ is injective if and only if $\ker\phi=$?  I think the $Ker\phi = \{x\in G:\phi(x)=1\}$  because you are looking at an identity then it is all reals $0$ does not work a because if $x=0, e'x=0$  but then i realize it is injective how does that affect the kernel?
79390
I have this written down in my notes, but cannot seem to find verification of it. How can I see this fact?   I believe there is a theorem, something along the lines of if a group G is isomorphic to H then, H is a group. Can anybody verify this or point me to the direction I am looking for? Thanks.
98497
So I tried looking everywhere but I'm really lost with this proof. I need to to show that:  Let $H$, $K$ be normal subgroups of a group $G$, satisfying $H\cap K = \{e\}$. Let $h\in H$, $k \in K$. (a) Prove that the commutator $w = hkh^{-1}k^{−1}$ satisfies $w \in H$ and $w \in K$. Then conclude that $w = e$ (b) Prove that $hk = kh$, for all $h \in H$, $k \in K$.
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.
104898
Let $f: G \to H$ be a homomorphism of groups. Assume that $a\in G$ and $\operatorname{ord}(a)=n$. Prove that the order of $f(a)$ is a divisor of $n$. I know that if $H$ is a subgroup of $G$, then $$|G|=|H|\cdot|\text{distinct cosets of $H$}|$$ so $|H|$ must be divisor of $|G|$, but I don't understand how to apply that logic to homomorphisms (or if that will even work). Any help is appreciated!
88728
Let $G$ be  a group and $H$ be a subgroup of $G$ of index $2$ .Then show that if $a,b\notin H$ then $ab\in H$.   $a,b\notin H\implies aH,bH\neq H\implies $ $aH=bH$  since $H$ has only two distinct left cosets in $G$.  $\implies a^{-1}b\in H$  How to show that $ab\in H$ from here?
36159
Is there a better (more direct or intuitive) proof for this proposition than I have come up with below? I am not sure whether it could be simiplified:  Let $G$ be a group with $H \leq G$. Then $K = \bigcap_{g \in G} gHg^{-1}$ is normal in $G$.  Let $a \in K$. Then $a \in gHg^{-1}$ for all $g \in G$. Therefore for all $g_1,g \in G$, $g_1ag_1^{-1} \in g_1gHg^{-1}g_1^{-1} = (g_1g)H(g_1g)^{-1}$ and so $g_1ag_1^{-1} \in K$ since $g_1g \in G$. Then $K$ is normal in $G$.